Class 12 Physics (Part 2) Chapter 25 Atom, Origin of Spectra : Bohr’s Theory of Hydrogen Atom

Key Concept: Mass number, Atomic number, Neutron number

c) 13
[Solution Description]
We know the formula for mass number is:
$A = Z + N$
Given: $A = 27$, $N = 14$.
Substituting the values:
$27 = Z + 14$
Solving for $Z$:
$Z = 27 - 14 = 13$.
Therefore, the atomic number is 13.

Key Concept: Discovery of Electron

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
J.J. Thomson indeed discovered electrons through his experiments with cathode rays, which showed that these rays consisted of negatively charged particles (electrons). Electrons are fundamental subatomic particles with a negative charge and are present in all atoms. The reason correctly explains why J.J. Thomson's discovery was significant—it identified the existence of negatively charged particles within atoms.

Key Concept: Lyman Series

d) $1215 \, \text{\AA}$
[Solution Description]
The first line in the Lyman series corresponds to the transition from $n = 2$ to $n = 1$. Using the formula:
$\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right)$
Simplifying,
$\frac{1}{\lambda} = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4}$
Therefore,
$\lambda = \frac{4}{3R} = \frac{4}{3 \times 1.097 \times 10^7} \approx 1215 \, \text{\AA}$

Key Concept: Orbit radii for different n

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The radius of the nth orbit for a hydrogen-like atom is given by $r_n = (0.53) \frac{n^2}{Z} \, \text{Å}$. For the hydrogen atom (Z = 1), the radius of the second orbit is:
$r_2^{\text{H}} = (0.53) \frac{2^2}{1} = 2.12 \, \text{Å}$
For the hydrogen-like atom with Z = 3:
$r_2^{\text{Z=3}} = (0.53) \frac{2^2}{3} = 0.706 \, \text{Å}$
Clearly, $0.706 < 2.12$, so Assertion (A) is true. The Reason (R) states that the radius is inversely proportional to Z, which is correct as per the formula. Additionally, the Reason correctly explains why the radius decreases with increasing Z. Hence, both are true, and the Reason is the correct explanation.

Key Concept: Distance of closest approach

b) It becomes half.
[Solution Description]
The distance of closest approach $r_0$ is given by:
$r_0 = \frac{1}{4\pi \epsilon_0} \cdot \frac{2Ze^2}{K}$
If $K$ is doubled, then $r_0$ becomes half of its original value because $r_0$ is inversely proportional to $K$.

Key Concept: Excitation potential, Energy levels, Wavelength calculation

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The Assertion states that the first excitation potential of hydrogen is 10.2 eV, which is correct as it matches the known energy difference between $n = 1$ (-13.6 eV) and $n = 2$ (-3.4 eV). The Reason explains that the emitted photon's wavelength can be calculated using $\lambda = \frac{hc}{\Delta E}$, which is also true. However, the Reason does not explain why the excitation potential is specifically 10.2 eV; it only describes how to calculate the wavelength of the emitted photon from this transition. Thus, both statements are true, but the Reason is not the correct explanation for the Assertion.

Key Concept: Rydberg Constant Variation

b) It decreases.
[Solution Description]
The Rydberg constant $R$ for a finite nuclear mass is given by:
$R = \frac{R_\infty}{1 + \frac{m}{M}}$, where $m$ is the electron mass and $M$ is the proton mass. Since $\frac{m}{M} > 0$, $R$ becomes smaller than $R_\infty$.

Key Concept: Energy-Level Diagram, Emission Spectra

b) $4.59 \times 10^{14} \, \text{Hz}$
[Solution Description]
The energy difference between the $-1.5 \, \text{eV}$ and $-3.4 \, \text{eV}$ levels is:
$\Delta E = -1.5 \, \text{eV} - (-3.4 \, \text{eV}) = 1.9 \, \text{eV}$
Using the relation $\Delta E = h\nu$, the frequency is:
$\nu = \frac{\Delta E}{h} = \frac{1.9}{4.14 \times 10^{-15}} \approx 4.59 \times 10^{14} \, \text{Hz}$

Key Concept: Excitation potential, Energy levels

b) 6.67 eV
[Solution Description]
First excitation potential corresponds to transition from ground state to first excited state (4.86 eV). Second excitation potential is for ground state to second excited state (6.67 eV). Therefore, minimum energy required to go directly from ground state to second excited state is equal to the second excitation potential = 6.67 eV.

Key Concept: Neutron Number for Isotopes

c) 8
[Solution Description]
The neutron number is given by:
$\text{Neutron Number} = A - Z$
Here, $A = 14$ and $Z = 6$. Substituting these values:
$\text{Neutron Number} = 14 - 6 = 8$
Thus, the neutron number for this carbon isotope is 8.

Key Concept: Identifying elements

b) It has 12 protons and 12 electrons
[Solution Description]
The atomic number ($Z$) uniquely identifies an element. If $Z = 12$, the element has 12 protons (and 12 electrons if neutral). Looking at the periodic table, the element with atomic number 12 is magnesium (\ce{Mg}).

Key Concept: Assertion-Reason

c) Assertion is true, but Reason is false.
[Solution Description]
The assertion states that most $\alpha$-particles passed through the gold foil undeflected, which is true as observed in the experiment. The reason claims this indicates a tiny, dense nucleus, but this conclusion arises from the few $\alpha$-particles deflected at large angles, not the undeflected ones. Thus, Assertion is true, but Reason misinterprets its significance.

Key Concept: Orbital period calculation

d) $12.24 \times 10^{-16} \, \text{s}$
[Solution Description]
The orbital period for the $n^{th}$ orbit is given by:
$T_n = T_1 \times n^3$
For $n=2$, substituting $T_1 = 1.53 \times 10^{-16} \, \text{s}$:
$T_2 = 1.53 \times 10^{-16} \times (2)^3 = 12.24 \times 10^{-16} \, \text{s}$

Key Concept: Distance of closest approach

b) It halves
[Solution Description]
The distance of closest approach ($r_0$) is given by:
$r_0 = \frac{1}{4\pi\epsilon_0} \cdot \frac{2Ze^2}{E_k}$
Here, $E_k$ is the kinetic energy of the $\alpha$-particle and $Z$ is the atomic number of the nucleus.
If $E_k$ is doubled ($E_k' = 2E_k$), the new distance of closest approach ($r_0'$) becomes:
$r_0' = \frac{1}{4\pi\epsilon_0} \cdot \frac{2Ze^2}{2E_k} = \frac{r_0}{2}$
Thus, $r_0$ is halved when $E_k$ is doubled.

Key Concept: No Energy Loss and Stability of Orbits

c) Electrons do not lose energy while revolving in stationary orbits.
[Solution Description]
According to Bohr's postulates:
1. Electrons revolve in stationary orbits without radiating energy.
2. They emit or absorb energy only when transitioning between orbits.
3. Classical electromagnetism predicts that accelerating charges radiate energy, but Bohr's theory contradicts this for stationary orbits.
The correct statement is that electrons do not lose energy while revolving in stationary orbits.

Key Concept: Fine structure, Bohr's limitations

c) Fine structure of spectral lines
[Solution Description]
Bohr's model successfully explains the spectral lines of hydrogen-like atoms but fails to explain the fine structure of spectral lines (splitting into multiple closely spaced lines), the relative intensities of spectral lines, the Zeeman effect (splitting under magnetic field), and the Stark effect (splitting under electric field). Among the options provided, the inability to explain the fine structure is a well-known limitation.

Key Concept: Wavelength Calculation

b) 122 nm
[Solution Description]
The energy difference between $n = 2$ and $n = 1$ is $\Delta E = E_2 - E_1 = (-3.4) - (-13.6) = 10.2 \text{ eV}$. Convert this energy to joules: $10.2 \times 1.6 \times 10^{-19} = 1.632 \times 10^{-18} \text{ J}$. Use the relation $\lambda = \frac{hc}{\Delta E}$:
$\lambda = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{1.632 \times 10^{-18}} \approx 1.22 \times 10^{-7} \text{ m} = 122 \text{ nm}.$

Key Concept: Ionisation potential, Energy levels

b) 54.4 eV
[Solution Description]
The energy of the ground state for a hydrogen-like ion is given by:
$E_1 = -13.6 Z^2 \text{ eV}$.
For Helium ion ($Z = 2$):
$E_1 = -13.6 \times (2)^2 = -13.6 \times 4 = -54.4 \text{ eV}$.
The ionisation energy is the difference between the ionised state ($E_\infty = 0$) and the ground state:
$E_{\text{ionisation}} = E_\infty - E_1 = 0 - (-54.4) = 54.4 \text{ eV}$.

Key Concept: Bohr's theory applicability, Multi-electron systems

d) Neutral helium ($He$)
[Solution Description]
Bohr's theory is only applicable to one-electron systems like hydrogen ($H$), singly-ionized helium ($He^+$), and doubly-ionized lithium ($Li^{2+}$). It fails for multi-electron atoms due to electron-electron interactions.

Key Concept: Radius of Permitted Orbits, Quantum Number Dependence

c) $r_6 = 4r_3$
[Solution Description]
Since the radius $r_n$ is proportional to $n^2$:
$r_n \propto n^2$
Therefore:
$\frac{r_6}{r_3} = \frac{6^2}{3^2} = \frac{36}{9} = 4$
Thus:
$r_6 = 4r_3$

Key Concept: Observations, Coulomb's Law

b) $\frac{d}{2}$
[Solution Description]
The distance of closest approach ($d$) occurs when the α-particle's kinetic energy is converted to electrostatic potential energy:
$E = \frac{k(2e)(Ze)}{d}$
where $k$ is Coulomb's constant, $2e$ is the charge of the α-particle, and $Ze$ is the nuclear charge. Rearranging:
$d = \frac{2kZe^2}{E}$
If $E$ is doubled ($E' = 2E$), the new distance $d'$ becomes:
$d' = \frac{2kZe^2}{2E} = \frac{kZe^2}{E} = \frac{d}{2}$

Key Concept: Scintillation Screen to Count Deflections

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description] The Assertion states that the scintillation screen was coated with zinc sulfide ($ZnS$) due to its light-emitting property when hit by $\alpha$-particles. This is true, as $ZnS$ indeed emits flashes of light (scintillations) upon impact with high-energy particles like $\alpha$-particles. The Reason explains that $ZnS$ fluoresces when interacting with such particles, which is also correct and directly justifies why $ZnS$ was used in the scintillation screen. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Rutherford's Atomic Model

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to Rutherford’s atomic model, electrons revolve around the nucleus in circular orbits. However, classical electrodynamics states that an accelerating charged particle (like an electron in circular motion) must emit radiation continuously. As the electron loses energy, it would spiral into the nucleus, making the atom unstable. This is a known drawback of Rutherford’s model. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Lyman series, shortest wavelength

d) 912 \, \text{\AA}
[Solution Description]
The shortest wavelength in the Lyman series corresponds to the transition from $n = \infty$ to $n = 1$. Using the Rydberg formula:
$\frac{1}{\lambda} = R \left( 1 - \frac{1}{\infty^2} \right) = R \left( 1 - 0 \right) = R$
Therefore,
$\lambda = \frac{1}{R} = \frac{1}{1.097 \times 10^7 \, \text{m}^{-1}} = 9.116 \times 10^{-8} \, \text{m} = 911.6 \, \text{\AA}$

Key Concept: Impact parameter and scattering angle

b) The scattering angle increases.
[Solution Description]
The relationship between impact parameter ($b$) and scattering angle ($\theta$) is given by:
$b = \frac{1}{4\pi \epsilon_0} \cdot \frac{Ze^2}{E} \cot\left(\frac{\theta}{2}\right)$
From this equation, as $b$ decreases, $\cot\left(\frac{\theta}{2}\right)$ also decreases. Since $\cot$ is a decreasing function, $\frac{\theta}{2}$ increases, meaning $\theta$ increases. Thus, the scattering angle increases as the impact parameter decreases.

Key Concept: Size of Nucleus

c) The nucleus is much smaller than the atom, with $r_n \approx 10^{-5} r_a$.
[Solution Description]
Rutherford concluded that the nucleus is extremely small compared to the overall size of the atom. The radius of the nucleus ($r_n$) is approximately $10^{-15}$ m, while the atomic radius ($r_a$) is about $10^{-10}$ m. Thus, $r_n \approx 10^{-5} r_a$.

Key Concept: Nuclear Size, Rutherford vs. Thomson Model

d) Rutherford’s model explains both undeflected trajectories and rare large-angle scattering.
[Solution Description]
Key differences:
1) Rutherford’s model predicts most α-particles passing undeflected (empty space) and a few repelled at large angles (tiny dense nucleus), while Thomson’s model predicts uniform small-angle deflections.
2) Only Rutherford’s model explains backscattering (>90°).

Key Concept: Wavelength Calculation

b) 2546 \AA
[Solution Description]
Given $\Delta E = 4.86 \, \text{eV}$, apply the formula:
$\lambda = \frac{12375}{4.86} = 2546.30 \, \text{\AA}$

Key Concept: Energy levels, quantization, transition probabilities

b) 0.512
[Solution Description]
Calculate $\Delta E_{5\to3}$ and $\Delta E_{3\to2}$ using the formula $E_n = -\frac{13.6}{n^2} \text{ eV}$.
For $n = 5$ to $n = 3$:
$E_5 = -\frac{13.6}{25} = -0.544 \text{ eV}, \quad E_3 = -\frac{13.6}{9} = -1.511 \text{ eV}$
$\Delta E_{5\to3} = E_3 - E_5 = -1.511 - (-0.544) = -0.967 \text{ eV}$
For $n = 3$ to $n = 2$:
$E_2 = -3.4 \text{ eV}, \quad \Delta E_{3\to2} = E_2 - E_3 = -3.4 - (-1.511) = -1.889 \text{ eV}$
The ratio is:
$\left|\frac{\Delta E_{5\to3}}{\Delta E_{3\to2}}\right| = \frac{0.967}{1.889} = 0.512$

Key Concept: Thomson's model, Limitations

c) Large-angle scattering of $\alpha$-particles
[Solution Description]
The plum pudding model proposed by Thomson suggested a uniform distribution of positive charge with electrons embedded within it. However, this model failed to explain the large-angle scattering of $\alpha$-particles observed in Rutherford's experiment, as such scattering would require a concentrated positive charge (nucleus), not a diffuse one.

Key Concept: Bohr's model limitations, de Broglie wavelength

c) $\frac{2\pi r}{n}$
[Solution Description]
To find the de Broglie wavelength, we use $\lambda = \frac{h}{p}$ where $p = m_e v$. From Bohr's quantization condition, $L = m_e v r = n\frac{h}{2\pi}$, so $v = \frac{nh}{2\pi m_e r}$. Substituting $v$ into $p$ gives $p = \frac{nh}{2\pi r}$. Thus, the de Broglie wavelength is $\lambda = \frac{h}{\frac{nh}{2\pi r}} = \frac{2\pi r}{n}$.
Therefore, the de Broglie wavelength is $\lambda = \frac{2\pi r}{n}$.

Key Concept: Wavelength Calculation

a) 589 nm
[Solution Description]
The energy of the emitted photon equals the first excitation energy ($E = 2.1$ eV). Using $E = \frac{hc}{\lambda}$, we solve for $\lambda$:
$\lambda = \frac{hc}{E} = \frac{(6.6 \times 10^{-34})(3 \times 10^8)}{2.1 \times 1.6 \times 10^{-19}} = 5.89 \times 10^{-7} \, \text{m} = 589 \, \text{nm}$

Key Concept: Absorption spectrum, Energy transitions

b) 2536 Å
[Solution Description]
Mercury atoms absorb specific energies corresponding to transitions from the ground state to higher energy levels. The problem states that wavelengths like 2536 Å and 1855 Å are absorbed.
Therefore, these wavelengths will be missing in the emerging light.

Key Concept: Energy of electron in hydrogen-like atom

b) $-3.4 \text{ eV}$
[Solution Description]
The energy of an electron in a hydrogen-like atom is given by $E = - \frac{Z^2 R h c}{n^2}$. For hydrogen, $Z = 1$ and ground state energy $E_1 = -13.6 \text{ eV}$. The energy for any level $n$ is $E_n = \frac{E_1/n^2}$.
Step 1: For $n = 2$, $E_2 = \frac{-13.6}{4} = -3.4 \text{ eV}$.

Key Concept: Energy levels, Bohr's model

d) 4
[Solution Description]
The energy levels of hydrogen are given by $E_n = -\frac{13.6 \text{ eV}}{n^2}$. The initial energy level ($n=1$) has $E_1 = -13.6$ eV. After absorbing 12.75 eV, the new energy is $-13.6 + 12.75 = -0.85$ eV. Solving $E_n = -0.85$ eV gives:
$-0.85 = -\frac{13.6}{n^2} \implies n^2 = \frac{13.6}{0.85} = 16 \implies n = 4.$

Key Concept: Relative orbit sizes, quantum numbers

b) 4
[Solution Description]
Radius for n=4: $r_4 = (0.53)(4)^2 = 8.48 Å$
Radius for n=2: $r_2 = (0.53)(2)^2 = 2.12 Å$
Decrease factor = $\frac{8.48}{2.12} = 4$

Key Concept: Excitation Potential

b) 4.86 eV
[Solution Description]
The excitation potential corresponds to the energy required to move an electron from the ground state to the first excited state. For mercury, this is 4.86 eV. Thus, the electron must have at least 4.86 eV of kinetic energy to cause this transition.

Key Concept: Velocity of Electron

c) It becomes half
[Solution Description]
The velocity of the electron in the $n^{th}$ orbit for a hydrogen-like atom is:
$v_n = \frac{Z e^2}{2 \epsilon_0 h n}$
Since $v_n$ is inversely proportional to $n$, when $n$ increases from 1 to 2, the velocity decreases:
$\frac{v_2}{v_1} = \frac{1}{2}$
Thus, the velocity becomes half.

Key Concept: Radius of Permitted Orbits

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to the formula for the radius of the $n$-th orbit of a hydrogen atom:
$r_n = (0.53) n^2 \, \text{Å}.$
For $n = 1$, the radius is $r_1 = 0.53 \, \text{Å}$.
For $n = 2$, the radius is $r_2 = (0.53) \times (2)^2 = 2.12 \, \text{Å}$, which is indeed four times $r_1$.
Also, the assertion correctly states that the radius is proportional to $n^2$ as given by the formula.
Therefore, both Assertion and Reason are true, and the Reason explains the Assertion correctly.

Key Concept: Assertion-Reason

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To verify the assertion and reason, we use the Rydberg formula for the Balmer series:
$\frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right)$
Substituting the value of the Rydberg constant $R = 1.097 \times 10^7 \, \text{m}^{-1}$:
$\lambda = \frac{36}{5 \times 1.097 \times 10^7} \approx 6.56 \times 10^{-7} \, \text{m} = 656 \, \text{nm}$
Thus, the Assertion (A) is true. The Reason (R) correctly identifies the quantum numbers ($n_1 = 2$, $n_2 = 3$) for the first line of the Balmer series, which is used to derive the wavelength. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Unstable Atom and Radiation Loss

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that classical electromagnetic theory predicts the collapse of an orbiting electron into the nucleus due to continuous energy radiation. This is correct because, according to classical theory, any accelerating charge (like an electron in circular motion) must radiate energy, leading to orbital decay. The reason explains why this happens—the accelerating electron radiates energy, causing it to lose energy and spiral inward. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Distance of closest approach

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that decreasing the initial kinetic energy ($K$) of the $\alpha$-particle increases the distance of closest approach ($r_0$). This can be verified using the given formula: $r_0 = \frac{1}{4\pi \epsilon_0} \cdot \frac{2Ze^2}{K}$. Here, $K$ is in the denominator, meaning if $K$ decreases, $r_0$ must increase. The reason provides the correct formula that explains why the assertion is true. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Hydrogen Spectrum Series

b) Balmer series
[Solution Description]
The Balmer series corresponds to transitions where electrons fall to the $n = 2$ level and emits light in the visible region. Other series like Lyman (UV), Paschen (IR), Brackett (IR), and Pfund (IR) lie outside the visible range.

Key Concept: Experimental confirmation

d) Davisson and Germer
[Solution Description]
In 1927, Davisson and Germer conducted an experiment where they observed electron diffraction patterns, which confirmed the wave-like behavior of electrons as predicted by de Broglie's hypothesis. This was a crucial verification of the dual nature of matter.

Key Concept: Nuclear Size Estimation

c) The nucleus is much smaller than the atom.
[Solution Description]
Most \$\alpha\$-particles passing through the gold foil without deflection indicates that the majority of the atom's volume is empty space. The nucleus, which caused large-angle deflections for a few particles, must be extremely small compared to the overall size of the atom. For gold, the nuclear radius is estimated to be around \$6.9 \times 10^{-15} \, \text{m}\$, while the atomic radius is on the order of \$10^{-10} \, \text{m}\$.

Key Concept: Zeeman Effect

b) It does not incorporate electron spin or angular momentum interactions
[Solution Description]
The Zeeman effect demonstrates that spectral lines split into multiple components in the presence of an external magnetic field. This phenomenon arises due to the interaction between the magnetic field and the orbital and spin angular momenta of the electron, which Bohr's model does not account for.

Key Concept: Absorption Spectrum

c) Most atoms are in the ground state, so transitions start from $n = 1$.
[Solution Description]
At ordinary temperatures, most hydrogen atoms remain in their ground state ($n = 1$). Absorption transitions can only occur if photons are absorbed to move electrons from $n = 1$ to higher energy levels, which corresponds to the Lyman series. Transitions starting from higher levels (e.g., $n = 2$, Balmer series) require atoms to already be in excited states, which are rare at ordinary temperatures.

Key Concept: Bohr's Atomic Model

b) $\frac{h}{\pi}$
[Solution Description]
The angular momentum of an electron in the $n^{th}$ orbit in Bohr's model is given by:
$mvr = n\frac{h}{2\pi}$
For the second orbit ($n = 2$), substitute $n = 2$ in the equation:
$mvr = 2\frac{h}{2\pi} = \frac{h}{\pi}$

Key Concept: Wave nature of electron, Relativistic effects

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that Bohr's model fails to explain the fine structure of the hydrogen spectrum due to its neglect of relativistic effects. This is true because Bohr's model assumes non-relativistic mechanics, whereas fine structure requires relativistic corrections. The Reason correctly identifies that fine structure arises from spin-orbit coupling and relativistic corrections. Further, the Reason accurately explains why the Assertion is true since both spin-orbit coupling (not included in Bohr's model) and relativistic effects contribute to fine structure splitting. Thus, both statements are true, and the Reason correctly explains the Assertion.

Key Concept: Emission Process

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that spectral lines emitted by an element are unique to that element and can be used for identification, which is true because each element has distinct energy levels. The reason provided is also correct as it explains why the emission spectra are unique—the energy states of atoms vary between elements. Moreover, the reason correctly justifies the assertion, as the uniqueness of energy states leads to unique spectral lines.
Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Bohr Model, Kinetic Energy

c) 1:4
[Solution Description]
Kinetic energy in Bohr model is given by:
$K_n = \frac{mZ^2e^4}{8\epsilon_0^2h^2n^2}$
Ratio of K$_2$/K$_1$:
$\frac{K_2}{K_1} = \frac{n_1^2}{n_2^2} = \frac{1^2}{2^2} = \frac{1}{4}$
Since Z is same for both states, it cancels out in the ratio.

Key Concept: Ionization Energy

c) 3.4 eV
[Solution Description]
The energy of the electron in the $n = 2$ state is given by:
$E_n = -\frac{13.6 \, \text{eV}}{n^2}$
For $n = 2$,
$E_2 = -\frac{13.6}{2^2} = -3.4 \, \text{eV}$
Ionization energy is the energy required to move the electron from its current state to infinity ($E_{\infty} = 0$). Thus,
$\text{Ionization Energy} = E_{\infty} - E_2 = 0 - (-3.4) = 3.4 \, \text{eV}$

Key Concept: Distance of Closest Approach

b) $2.5 \times 10^{-14}$ m
[Solution Description]
Using the formula for the distance of closest approach:
$r_0 = \frac{1}{4 \pi \epsilon_0} \cdot \frac{2Ze^2}{K}$
Substituting the given values:
$r_0 = 9 \times 10^9 \cdot \frac{2 \times 79 \times (1.6 \times 10^{-19})^2}{1.48 \times 10^{-12}}$
Calculating step-by-step:
$r_0 = 9 \times 10^9 \cdot \frac{2 \times 79 \times 2.56 \times 10^{-38}}{1.48 \times 10^{-12}} = 9 \times 10^9 \cdot \frac{404.48 \times 10^{-38}}{1.48 \times 10^{-12}} = 9 \times 10^9 \cdot 273.3 \times 10^{-26} = 2.46 \times 10^{-14} \, \text{m}$
So, the distance of closest approach is approximately $2.5 \times 10^{-14}$ m.

Key Concept: De Broglie wavelength, Stationary orbits

c) $0.332 \, \text{nm}$
[Solution Description]
The de Broglie wavelength ($\lambda$) of the electron is given by:
$\lambda = \frac{h}{mv}$
Substituting the given values:
$\lambda = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \times 2.19 \times 10^6} = 3.32 \times 10^{-10} \, \text{m} = 0.332 \, \text{nm}$

Key Concept: Radii of Bohr's Orbits

c) $4.77 \times 10^{-10} \, \text{m}$
[Solution Description]
The radius of the $n^{\text{th}}$ orbit is given by:
$r_n = n^2 r_1$
For $n = 3$:
$r_3 = 9 \times 5.3 \times 10^{-11} \, \text{m} = 4.77 \times 10^{-10} \, \text{m}$

Key Concept: Time of Collapse

b) $10^{-8}$ seconds
[Solution Description]
Based on classical electromagnetic theory, an electron orbiting the nucleus would lose energy rapidly due to radiation emission and collapse into the nucleus in about $10^{-8}$ seconds. This contradicts the observed stability of atoms.

Key Concept: Discovery of Electron

c) J.J. Thomson
[Solution Description]
J.J. Thomson discovered the electron in 1897, as mentioned in the syllabus. This discovery was a key milestone in atomic theory.

Key Concept: Basis for spectral identification of elements

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Assertion (A) is true because the energy levels in an atom are quantized and unique for each element, leading to characteristic emission/absorption spectra.
Reason (R) is also true as it correctly explains the relationship between energy difference ($\Delta E$) and wavelength ($\lambda$) using Planck's equation $\Delta E = \frac{hc}{\lambda}$.
Since both are true and R explains A, the correct answer is a).

Key Concept: Second excitation potential of mercury

a) 186 \, \text{nm}
[Solution Description]
The second excitation potential corresponds to $\Delta E = 6.67 \, \text{eV} = 6.67 \times 1.6 \times 10^{-19} \, \text{J}$.
The wavelength $\lambda$ is calculated as:
$\lambda = \frac{hc}{\Delta E} = \frac{(6.6 \times 10^{-34})(3.0 \times 10^8)}{6.67 \times 1.6 \times 10^{-19}} = \frac{1.98 \times 10^{-25}}{1.067 \times 10^{-18}} \approx 1.86 \times 10^{-7} \, \text{m} = 186 \, \text{nm}.$

Key Concept: Observations from the Experiment

c) They passed straight through without deflection.
[Solution Description]
The majority of α-particles passed straight through the gold foil without any deflection. This indicates that most of the atom is empty space, as very few α-particles encountered a force strong enough to deflect them.

Key Concept: Experimental setup

b) Crossed electric and magnetic fields
[Solution Description]
The crossed electric and magnetic fields in Thomson's apparatus were essential as they allowed balancing of forces (Lorentz force law). By adjusting these fields until the beam was undeflected, he could calculate the velocity and then the charge-to-mass ratio of electrons.

Key Concept: Absorption Spectrum and Energy Levels

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the absorption spectrum of hydrogen at ordinary temperatures shows only the Lyman series, which is true because at normal temperatures, almost all hydrogen atoms remain in their lowest energy level ($n = 1$). Thus, absorption transitions can only start from the ground state, leading to the Lyman series in the absorption spectrum. The reason correctly explains why the Lyman series is observed by stating that most hydrogen atoms are in the ground state at ordinary temperatures, making it the starting point for absorption transitions.

Key Concept: Bohr's Postulate on Stationary Orbits

c) Stationary orbits
[Solution Description]
According to Bohr's third postulate, electrons revolve in stable or stationary orbits without radiating energy despite their acceleration towards the nucleus. These orbits are called stationary orbits.

Key Concept: Bohr's Frequency Condition

d) The frequency is directly proportional to the energy difference
[Solution Description]
Bohr's frequency condition states that when an electron transitions from a higher energy level $E_2$ to a lower energy level $E_1$, the frequency $v$ of the emitted photon is given by:
$hv = E_2 - E_1$
Thus, the frequency of the emitted radiation is directly proportional to the energy difference between the two orbits.

Key Concept: Radius of Permitted Orbits, Atomic Number Dependence

b) 2
[Solution Description]
The formula for the radius of the first orbit ($n = 1$) is:
$r_1 = (0.53) \cdot \frac{1}{Z} \, \text{\AA}$
Given $r_1 = 0.265 \, \text{\AA}$:
$0.265 = \frac{0.53}{Z}$
Solving for $Z$:
$Z = \frac{0.53}{0.265} = 2$

Key Concept: Distance of closest approach, Nuclear charge

b) Increases to four times original value
[Solution Description]
The distance of closest approach formula is:
$r_0 = \frac{1}{4\pi\epsilon_0} \cdot \frac{2Ze^2}{K}$
For initial case: \$r_{01} = \frac{1}{4\pi\epsilon_0} \cdot \frac{2Ze^2}{K}\$
New conditions:
- Energy becomes \$2K\$
- Atomic number becomes \$4Z\$
New distance: \$r_{02} = \frac{1}{4\pi\epsilon_0} \cdot \frac{2(4Z)e^2}{2K} = \frac{1}{4\pi\epsilon_0} \cdot \frac{8Ze^2}{2K} = 4 \cdot \frac{1}{4\pi\epsilon_0} \cdot \frac{2Ze^2}{K} = 4r_{01}\$
Therefore, the distance increases by factor of 4.

Key Concept: Energy Levels

b) 1.89 eV
[Solution Description]
The energy in the second excited state ($n = 3$) is $E_3 = -1.51 \, \text{eV}$.
The energy in the first excited state ($n = 2$) is $E_2 = -3.4 \, \text{eV}$.
The energy of the emitted photon is:
$\Delta E = E_3 - E_2 = -1.51 \, \text{eV} - (-3.4 \, \text{eV}) = 1.89 \, \text{eV}.$

Key Concept: Observations from Rutherford's α-Particle Scattering Experiment

c) The atom is mostly hollow.
[Solution Description]
The experiment showed that most α-particles passed through the gold foil without deflection, meaning the majority of the atom is empty space. Thus, the correct conclusion is that the atom is mostly hollow.

Key Concept: Bohr's Quantization, De Broglie Wavelength

b) 4:1
[Solution Description]
Step 1: Express momentum in terms of angular momentum.
$L = pr = n\frac{h}{2\pi} \Rightarrow p = \frac{nh}{2\pi r}$.
Step 2: Substitute into de Broglie wavelength.
$\lambda = \frac{h}{p} = \frac{h(2\pi r)}{nh} = \frac{2\pi r}{n}$.
Step 3: Since $r \propto n^2$, $\lambda \propto \frac{n^2}{n} = n$.
Thus, $\frac{\lambda_1}{\lambda_4} = \frac{1}{4}$.

Key Concept: Ionisation energy, Discrete energy levels

b) 3
[Solution Description]
First calculate the highest energy level reached: Total energy = $-13.6 + 15.0 = 1.4$ eV. Using $E_n = \frac{-13.6}{n^2}$, solving $-1.4 = \frac{-13.6}{n^2}$ gives $n \approx 3.12$, so maximum $n = 3$. Possible transitions are: 3→2, 3→1, and 2→1, giving 3 different photon energies.

Key Concept: Numerical Applications - Balmer Series Wavelength Calculation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that the wavelength of the first line in the Balmer series is 6563 Å. This is correct because for $n = 3$, substituting in the formula $\frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{5R}{36}$ and using $R = 1.097 \times 10^7 \, \text{m}^{-1}$, we get $\lambda = \frac{36}{5 \times 1.097 \times 10^7} = 6.563 \times 10^{-7} \, \text{m} = 6563 \, \text{\AA}$. The Reason provides the correct formula for the wavelength in the Balmer series, which explains why the Assertion is true.

Key Concept: Radii of permitted orbits

c) 2.12 \AA
[Solution Description]
For hydrogen atom, the radius of the $n$-th orbit is given by:
$r_n = (0.53) n^2 \, \text{Å}$
Substitute $n = 2$:
$r_2 = 0.53 \times (2)^2 = 2.12 \, \text{Å}$
Therefore, the radius of the second orbit is $2.12 \, \text{Å}$.

Key Concept: Classical theory of radiation, Stability of atoms

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
Classical electromagnetic theory predicts that an accelerating electron (such as one moving in a circular orbit around the nucleus) must emit radiation continuously. This emission would cause the electron to lose energy and spiral inward, leading to collapse. However, experimental observations show that atoms are stable and do not collapse, with electrons occupying fixed orbits without radiating energy. Additionally, atomic spectra consist of discrete lines rather than a continuous spectrum, which is another contradiction to the classical prediction. The Assertion correctly states the classical prediction, and the Reason correctly identifies the experimental observation that contradicts this prediction, but the Reason does not explain why the Assertion happens—it merely highlights a discrepancy. Thus, both are true, but the Reason does not explain the Assertion.

Key Concept:

b) To prevent unwanted scattering by air molecules
[Solution Description]
The vacuum chamber prevented \$\alpha\$-particles from colliding with air molecules, ensuring that observed scattering was solely due to interactions with the gold atoms' nuclei.

Key Concept: Transitions, Spectral lines

b) 10
[Solution Description]
The number of possible spectral lines when an electron jumps from any higher energy level to lower levels is given by the number of possible transitions. For $n=5$ to $n=1$, it can jump directly or via intermediate levels. The number of possible transitions is:
$\frac{n(n-1)}{2} = \frac{5 \times 4}{2} = 10$
Thus, there are 10 possible spectral lines.

Key Concept: Ion Neutrality

c) $+1$
[Solution Description]
A neutral lithium atom has 3 protons and 3 electrons. When it loses 1 electron, there are 3 protons but only 2 electrons left. The net charge is calculated as:
Net charge = Total proton charge + Total electron charge
Net charge = $(+3e) + (-2e) = +1e$.
Hence, the net charge is $+1$.

Key Concept: Atomic Number and Mass Number

a) 6
[Solution Description]
The neutron number ($N$) is given by $N = A - Z$, where $A$ is the mass number and $Z$ is the atomic number. For this element, $A = 12$ and $Z = 6$. Substituting these values, we get $N = 12 - 6 = 6$.

Key Concept: Charge of Electron

c) Negative
[Solution Description]
According to Thomson's discovery, the electron is a negatively charged particle. The atom as a whole remains neutral due to the balance of positive and negative charges.

Key Concept: Bohr's theory and line spectra

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that Bohr's model explains emission of radiation during electron transitions between quantized energy levels, which is correct. The reason provides the formula $\Delta E = h \nu$, which relates the energy difference to the emitted radiation's frequency. This formula correctly explains why discrete spectral lines are observed, as it connects quantized energy levels with specific radiation frequencies. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Energy levels and transitions

b) 4853 Å
[Solution Description]
Given: $\Delta E = 2.55$ eV. Using the formula for wavelength,
$\lambda = \frac{12375}{2.55} = 4853 \text{ Å}$
The wavelength of the emitted photon is 4853 Å.

Key Concept: Hydrogen spectrum

b) 10.2 eV
[Solution Description]
The first excitation energy for hydrogen is the energy required to move an electron from $n=1$ to $n=2$. Using $E_n = -\frac{13.6}{n^2}$ eV:
$\Delta E = E_2 - E_1 = \left(-\frac{13.6}{4}\right) - (-13.6) = -3.4 + 13.6 = 10.2 \text{ eV}.$

Key Concept: Balmer Series

c) $3646 \, \text{\AA}$
[Solution Description]
The shortest wavelength in the Balmer series corresponds to the transition where $n \rightarrow \infty$. Using the formula:
$\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{\infty} \right) = \frac{R}{4}$
Therefore,
$\lambda = \frac{4}{R} = \frac{4}{1.097 \times 10^7} \approx 3646 \, \text{\AA}$

Key Concept: Hydrogen Spectral Series

b) Balmer series
[Solution Description]
The Balmer series corresponds to transitions where the electron falls to the $n = 2$ level, emitting photons in the visible region of the electromagnetic spectrum.

Key Concept: Transition between energy levels

c) The atom emits energy
[Solution Description]
According to Bohr's model, when an electron transitions from a higher energy level ($n = 2$) to a lower energy level ($n = 1$), it emits energy equal to the difference in energy between the two levels. This emitted energy corresponds to a photon of specific frequency given by $\Delta E = h\nu$.

Key Concept: Scattering angle dependence, Number of scattered particles

a) About N/14
[Solution Description]
The number of scattered particles follows $N \propto \frac{1}{\sin^4(\theta/2)}$.
Step-by-step calculation:
1. For $\theta_1 = 30^\circ$: $\sin(15^\circ) \approx 0.2588$, so $N_1 \propto \frac{1}{(0.2588)^4} \approx 226$
2. For $\theta_2 = 60^\circ$: $\sin(30^\circ) = 0.5$, so $N_2 \propto \frac{1}{(0.5)^4} = 16$
3. Ratio: $\frac{N_2}{N_1} = \frac{226}{16} \approx 14$ times less
4. Therefore, $N_2 \approx \frac{N}{14}$

Key Concept: Excitation energy, Ionisation potential

b) $n = 3$
[Solution Description]
Ground state energy of hydrogen atom = $-13.6$ eV. The energy after absorption is $-13.6 + 12.1 = -1.5$ eV. The energy of nth level in hydrogen atom is given by $E_n = \frac{-13.6}{n^2}$ eV. Solving $-1.5 = \frac{-13.6}{n^2}$, we get $n^2 = \frac{13.6}{1.5} \approx 9.07$. Taking square root gives $n \approx 3$.

Key Concept: Limitations of Bohr’s Theory

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The fine structure of spectral lines arises due to the interaction between the electron's spin and its orbital motion, which was not considered in Bohr's theory. Therefore, while the assertion is true as it correctly states one of the limitations of Bohr's theory, the reason is also true and provides the correct explanation for this limitation.

Key Concept: Atomic Number and Nucleus Composition

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the atomic number of lithium is 3 because its nucleus contains 3 protons, which is true as per the given syllabus. The reason states that the atomic number of an element is equal to the number of protons in its nucleus, which is also true and correctly explains why lithium has an atomic number of 3. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Stationary Orbits

c) Stationary orbits forbid energy loss due to radiation
[Solution Description]
Bohr's third postulate explains that electrons in stationary orbits do not radiate energy despite their acceleration because these orbits are stable states. The electron remains in such an orbit without losing energy unless it transitions between orbits.

Key Concept: Distance of Closest Approach

c) \$r_0\$ decreases by a factor of 4.
[Solution Description]
The distance of closest approach is given by: $r_0 = \frac{1}{4\pi \epsilon_0} \cdot \frac{2Ze^2}{K}$
Here, \$r_0 \propto \frac{1}{K}\$. If the kinetic energy \$K\$ is increased by a factor of 4, then \$r_0\$ decreases by a factor of 4 because it is inversely proportional to \$K\$.

Key Concept: Neutron Number Calculation

d) 4
[Solution Description]
The neutron number is calculated using the formula $\text{Neutron Number} = A - Z$. Given $Z = 3$ and $A = 7$, substituting these values gives $\text{Neutron Number} = 7 - 3 = 4$.

Key Concept: Impact Parameter and Scattering Angle

b) $1.2 \times 10^{-14}$ m
[Solution Description]
The impact parameter ($b$) is related to the scattering angle ($\theta$) by:
$b = \frac{1}{4\pi \epsilon_0} \cdot \frac{Ze^2}{E} \cot\left(\frac{\theta}{2}\right)$
Given $\theta = 90^\circ$, $\cot(45^\circ) = 1$. Substituting the values:
$b = 9 \times 10^9 \cdot \frac{79 \times (1.6 \times 10^{-19})^2}{1.48 \times 10^{-12}} \cdot 1$
$b = 9 \times 10^9 \cdot \frac{79 \times 2.56 \times 10^{-38}}{1.48 \times 10^{-12}}$
$b = 9 \times 10^9 \cdot 136.65 \times 10^{-26}$
$b = 1.23 \times 10^{-14} \, \text{m}$

Key Concept: Excitation potential, energy levels

c) $n=4$
[Solution Description]
The energy of the absorbed photon is calculated using the formula:
$E = \frac{hc}{\lambda}$.
Plugging in the values:
$E = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{97.5 \times 10^{-9} \, \text{m}}$.
This gives $E \approx 2.04 \times 10^{-18} \, \text{J}$.
Convert this energy to eV:
$E = \frac{2.04 \times 10^{-18}}{1.6 \times 10^{-19}} \approx 12.75 \, \text{eV}$.
The ground state energy of hydrogen is $-13.6 \, \text{eV}$. The energy after absorption is $-13.6 + 12.75 = -0.85 \, \text{eV}$.
The energy levels of hydrogen are given by $E_n = -\frac{13.6}{n^2} \, \text{eV}$.
Solving for $n$ when $E_n = -0.85 \, \text{eV}$:
$-0.85 = -\frac{13.6}{n^2}$,
$\Rightarrow n^2 = \frac{13.6}{0.85} = 16$,
$\Rightarrow n = 4$.
Thus, the electron transitions to the $n=4$ state.

Key Concept: Atomic Number and Mass Number

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the atomic number determines an element's identity, which is correct because the atomic number specifies the number of protons, defining the element's unique properties. The reason states that the atomic number equals the number of protons, which is also true and directly explains why the atomic number defines the element. Therefore, both are true, and the reason correctly explains the assertion.

Key Concept: Spectral Series

b) Lyman series
[Solution Description]
The Lyman series corresponds to transitions where the electron falls to the $n = 1$ level, emitting ultraviolet light. Other series like Balmer (visible), Paschen (infrared), Brackett (infrared), and Pfund (infrared) do not lie entirely in the ultraviolet region.

Key Concept: Bohr's Quantization of Angular Momentum

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to Bohr’s theory, the angular momentum of an electron in the $n^{th}$ orbit is quantized and is given by the formula $mvr = n \frac{h}{2\pi}$. Here, $h$ is Planck’s constant, and $n$ is the principal quantum number. Thus, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Impact Parameter

b) Inversely proportional to $\cot(\theta/2)$
[Solution Description]
The impact parameter $b$ is related to the scattering angle $\theta$ by:
$b = \frac{Ze^2 \cot(\theta/2)}{4\pi \epsilon_0 E_k}$
As $\theta$ increases, $\cot(\theta/2)$ decreases, leading to a smaller $b$. Therefore, $b$ is inversely proportional to $\cot(\theta/2)$.

Key Concept: Spectral series, Bohr's theory

c) 486 nm
[Solution Description]
The wavelength $\lambda$ of the emitted photon can be calculated using the Rydberg formula for the Balmer series:
$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
Here, $n_1 = 2$ and $n_2 = 4$. Substituting these values:
$\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{16} \right)$
$\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{3}{16} \right) = 2.056875 \times 10^6 \, \text{m}^{-1}$
$\lambda = \frac{1}{2.056875 \times 10^6} \approx 486 \, \text{nm}$
This corresponds to the blue-green region of the visible spectrum.

Key Concept: Thomson’s Plum-Pudding Model – failed to explain scattering

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Thomson’s plum-pudding model proposed that atoms consist of electrons embedded in a uniformly distributed positively charged sphere. However, this model could not account for the large-angle scattering observed in the $\alpha$-particle experiments because:
1) Large-angle scattering requires a concentrated positive charge (nucleus) to provide the strong repulsive force needed, but Thomson’s model lacked such concentration.
2) The uniform positive charge distribution would only result in minor deflections, as the repulsion would not be strong enough to cause significant scattering.
Therefore, the Assertion (A) is correct as the model indeed failed to explain the large-angle scattering. The Reason (R) is also true and correctly explains why the model failed—it assumed a uniform positive charge distribution that could not produce the necessary repulsive forces. Hence, both are true, and R is the correct explanation of A.

Key Concept: Line Spectra vs. Continuous Spectrum

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion is true because classical electromagnetism predicts that an accelerating charged particle (like an electron in orbit) would emit radiation continuously, leading to a continuous spectrum. The Reason is also true because, according to classical theory, this continuous emission of radiation would cause the electron to lose energy and spiral into the nucleus. Moreover, the Reason correctly explains why classical physics predicts a continuous spectrum, as the continuous energy loss implies a continuous range of emitted frequencies. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Emission of Light

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that an electron emits a photon only when it transitions from a higher energy level to a lower one, which is true according to atomic physics. The Reason explains that the emitted photon's energy equals the energy difference between the levels, which is also correct and directly explains the Assertion.

Key Concept: Bohr's Theory, Astrophysical Applications

b) Moving away from Earth at 695 km/s
[Solution Description]
1) The observed wavelength shift indicates Doppler effect:
$$ \frac{\Delta \lambda}{\lambda_0} = \frac{v}{c} $$
2) Given:
$$ \lambda_0 = 656.28 \text{nm} $$
$$ \lambda_{obs} = 657.8 \text{nm} $$
$$ \Delta \lambda = 657.8 - 656.28 = 1.52 \text{nm} $$
3) Calculate velocity:
$$ v = c \times \frac{\Delta \lambda}{\lambda_0} $$
$$ v = 3 \times 10^8 \times \frac{1.52}{656.28} $$
$$ v \approx 694,980 \text{m/s} \approx 695 \text{km/s} $$
4) Since wavelength increased (redshift), the star is moving away from Earth.

Key Concept: Energy difference and photon wavelength

b) 1024 Å
[Solution Description]
The energy levels of hydrogen are given by $E_n = -\frac{13.6}{n^2}$ eV. For $n = 3$, $E_3 = -\frac{13.6}{9} = -1.511$ eV. For $n = 1$, $E_1 = -13.6$ eV. The energy difference is $\Delta E = E_3 - E_1 = -1.511 - (-13.6) = 12.089$ eV. Using $\lambda = \frac{12375}{\Delta E}$, we get $\lambda = \frac{12375}{12.089} = 1023.7$ Å.

Key Concept: Rutherford's Alpha Scattering Experiment

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that very few $\alpha$-particles were deflected at angles greater than $90^\circ$ in Rutherford's gold foil experiment. This is true because most $\alpha$-particles passed straight through the atom, indicating the presence of empty space. However, a very small fraction was deflected at large angles, which implies they came very close to a small, dense, positively charged nucleus. The reason explains why this deflection occurs: due to the small size and high charge density of the nucleus, only a few $\alpha$-particles get close enough to experience the strong Coulomb repulsion needed for such large-angle deflections. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Hydrogen Spectrum, Energy Transitions

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The shortest wavelength in any spectral series corresponds to the transition where the electron falls from $n = \infty$ to the lower level ($n_1$) of that series. For the Balmer series, $n_1 = 2$.
The energy of the emitted photon is given by:
$E = Rhc\left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
When $n_2 \to \infty$, the second term becomes negligible:
$E_{\text{max}} = \frac{Rhc}{4}$
Since $E = h\nu = \frac{hc}{\lambda}$, the shortest wavelength corresponds to the highest energy (maximum frequency). Thus, the assertion is true.
The reason states that this transition has the maximum energy difference for Balmer series transitions, which is correct because as $n_2$ increases, the energy difference decreases, approaching $\frac{Rhc}{4}$ asymptotically. Therefore, the reason correctly explains the assertion.
Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Key Findings from Rutherford's Experiment

d) The mass and positive charge are concentrated in the nucleus.
[Solution Description]
Rutherford's experiment showed that the mass and positive charge are concentrated in a tiny nucleus, contradicting Thomson’s model where charge was uniformly distributed like "plums" in a "pudding."

Key Concept: Thomson's Model

b) Electrons are uniformly distributed in a positive charge cloud
[Solution Description]
Thomson's model describes the atom as a positively charged sphere with electrons embedded in it, like plums in a pudding. The electrons are in stable equilibrium, and the mass is distributed continuously throughout the atom.

Key Concept: Hydrogen Spectrum Conditions, Temperature Effect

b) The high temperature in the sun excites many hydrogen atoms to $n = 2$, enabling Balmer series absorption.
[Solution Description]
At ordinary temperatures, most hydrogen atoms are in the ground state ($n = 1$), so only Lyman series transitions (starting from $n = 1$) occur in the absorption spectrum. In the sun, the high temperature excites many hydrogen atoms to higher energy levels like $n = 2$, enabling Balmer series absorption transitions (starting from $n = 2$). Thus, both Lyman and Balmer series appear in the solar absorption spectrum.

Key Concept: Ionisation Energy Calculation

b) 1.51 eV
[Solution Description]
The energy of the electron in the ground state ($n = 1$) is $-13.6 \text{ eV}$. In the $n = 3$ state, the energy is given by $E_3 = -\frac{13.6}{3^2} = -1.51 \text{ eV}$. To remove the electron, we need to supply energy equal to the binding energy at $n = 3$, which is $1.51 \text{ eV}$ (since the electron must be raised to $E = 0$). Thus, the ionisation energy in this case is $1.51 \text{ eV}$.

Key Concept: De Broglie Hypothesis

c) $\lambda = \frac{h}{p}$
[Solution Description]
De-Broglie's hypothesis states that the wavelength $\lambda$ associated with an electron's momentum $p$ is given by $\lambda = \frac{h}{p}$, where $h$ is Planck's constant. This wave-like behavior is ignored in Bohr's model.

Key Concept: Alpha-particle scattering experiment

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that most alpha-particles pass straight through the gold foil without deflection, which is true as observed in Rutherford's experiment. The reason given is that the atom is mostly hollow with a small, dense nucleus, which explains why most alpha-particles are not deflected. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Discovery of Electron

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion is true because J.J. Thomson indeed used cathode ray tubes in his experiments, leading to the discovery of electrons. The reason is also correct as cathode rays were identified to be composed of negatively charged particles (electrons), which have a much smaller mass than hydrogen atoms. Moreover, the reason correctly explains why Thomson concluded the existence of electrons from his observations.

Key Concept: Transition wavelength

b) $n = 4$ to $n = 2$
[Solution Description]
First, find the energy difference: $\Delta E = \frac{12375}{4861} = 2.546$ eV. Now, check the possible transitions. For $n = 4$ to $n = 2$, $\Delta E = E_4 - E_2 = -0.85 - (-3.40) = 2.55$ eV, which matches closely with the given $\Delta E$.

Key Concept: Rutherford Scattering Law

c) $N \propto \frac{1}{\sin^4(\theta/2)}$
[Solution Description]
Rutherford's scattering law states that the number of scintillations is inversely proportional to the fourth power of the sine of half the scattering angle, given by:
$N \propto \frac{1}{\sin^4(\theta/2)}$

Key Concept: $\alpha$-Particle Scattering Experiment

c) A small fraction of $\alpha$-particles were scattered at large angles
[Solution Description]
Most $\alpha$-particles passed through the gold foil undeflected, but a very few were deflected at large angles (even backscattered). This suggested that the positive charge and mass are concentrated in a small region (nucleus).

Key Concept: Ionisation Potential

d) 13.6 V
[Solution Description]
The ionisation energy of a hydrogen atom is the energy required to remove an electron from the ground state ($n = 1$) to infinity ($n = \infty$).
This is equal to $13.6 \, \text{eV}$.
Since $1 \, \text{eV}$ corresponds to $1 \, \text{V}$ for a single electron, the ionisation potential is equal to the ionisation energy.
Therefore, ionisation potential = $13.6 \, \text{V}$.

Key Concept: Energy States and Quantum Transitions

b) $-3.4 \text{ eV}$
[Solution Description]
The energy of an electron in the nth orbit is given by:
$E_n = -\frac{13.6}{n^2} \text{ eV}$
For the second orbit ($n = 2$),
$E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \text{ eV}$
Hence, the correct energy is $-3.4 \text{ eV}$.

Key Concept: Quantization Condition, Energy Emission and Absorption

b) 2
[Solution Description]
Angular momentum in Bohr's model is quantized as $L = n\frac{h}{2\pi}$. For the second orbit ($n=2$) and first orbit ($n=1$), we calculate their angular momenta:
$L_{\text{initial}} = 2 \times \frac{h}{2\pi}, \quad L_{\text{final}} = 1 \times \frac{h}{2\pi}$
The ratio is $\frac{L_{\text{initial}}}{L_{\text{final}}} = \frac{2}{1} = 2$.

Key Concept: Energy Levels and Photon Wavelength

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a hydrogen atom, the energy levels are given by $E_n = -\frac{13.6}{n^2} \, \text{eV}$. The energy difference for the transition from $n=4$ to $n=2$ is $\Delta E = E_4 - E_2 = -\frac{13.6}{16} - \left(-\frac{13.6}{4}\right) = 2.55 \, \text{eV}$. Using the formula $\lambda = \frac{12375}{\Delta E}$, we get $\lambda = \frac{12375}{2.55} = 4853 \, \text{Å}$. Thus, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Excitation Energy

b) 12.09 eV
[Solution Description]
The energy of the electron in the ground state of the hydrogen atom is $E_1 = -13.6 \, \text{eV}$.
The energy in the second excited state ($n = 3$) is $E_3 = -1.51 \, \text{eV}$.
The excitation energy is calculated as:
$\Delta E = E_3 - E_1 = -1.51 \, \text{eV} - (-13.6 \, \text{eV}) = 12.09 \, \text{eV}.$

Key Concept: Hydrogen Spectral Series

d) Balmer series, 486.1 nm
[Solution Description]
The transition from $n = 4$ to $n = 2$ lies in the Balmer series because the final energy level is $n_1 = 2$. To find the wavelength $\lambda$, we use the formula for the Balmer series:
$\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right)$
Substituting $R = 1.097 \times 10^7$ m$^{-1}$,
$\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{16} \right) = 1.097 \times 10^7 \left( \frac{3}{16} \right) = 2.056875 \times 10^6 \, \text{m}^{-1}$
Therefore, $\lambda = \frac{1}{2.056875 \times 10^6} = 486.13 \, \text{nm}$.

Key Concept: Electron velocity \& period in Bohr's model

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
From Bohr's model, the speed of an electron in a hydrogen-like atom is given by:
$v = \frac{Ze^2}{2 \epsilon_0 h n}$
For hydrogen ($Z = 1$) and $n = 2$, the speed becomes:
$v_2 = \frac{e^2}{2 \epsilon_0 h \times 2} = \frac{v_1}{2} = \frac{2.18 \times 10^6}{2} = 1.09 \times 10^6$ m/s
Thus, Assertion (A) is true.
The Reason (R) states that the speed is inversely proportional to $n$, which matches the formula $v \propto \frac{1}{n}$. Hence, both (A) and (R) are true, and (R) correctly explains (A).

Key Concept: Scintillation screen, Angular distribution of $\alpha$-particles

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion states that the number of scintillations at $180^\circ$ is higher than at smaller angles. However, this contradicts the Rutherford scattering formula, which shows that the number of $\alpha$-particles scattered ($N$) is inversely proportional to $\sin^4(\theta/2)$.
Let's evaluate for $\theta = 180^\circ$ and $\theta = 90^\circ$:
For $\theta = 180^\circ$: $\sin(180^\circ/2) = \sin(90^\circ) = 1 \Rightarrow N \propto 1/1^4 = 1$.
For $\theta = 90^\circ$: $\sin(45^\circ) \approx 0.707 \Rightarrow N \propto 1/(0.707)^4 \approx 4$.
Thus, fewer particles are scattered at larger angles like $180^\circ$ compared to smaller angles like $90^\circ$. The assertion is false because it claims the opposite behavior. The reason correctly states the relationship from the Rutherford formula but misinterprets its implications when explaining the assertion. Therefore, the assertion is false while the reason is true.

Key Concept: Emission and Absorption Spectra

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Lyman series is indeed observed in the ultraviolet region because the energy differences involved in transitions to the ground state ($n = 1$) are large, resulting in photons with high energy and short wavelengths (UV). The reason correctly explains the assertion since the Lyman series corresponds to transitions ending at $n = 1$, which emit UV radiation. Both the assertion and reason are true, and the reason is the correct explanation.

Key Concept: Rutherford's Alpha-Particle Scattering Experiment

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that very few alpha-particles were scattered at angles greater than 90°, which indicates that the positive charge in an atom is concentrated in a very small space called the nucleus. This is true based on Rutherford's experiment, where only about 1 in 20,000 alpha-particles were deflected at large angles. The reason states that most alpha-particles passed straight through the gold foil because most of the space inside the atom is empty. This is also true, as it explains why the majority of particles were undeflected. Furthermore, the reason correctly explains the assertion: the concentrated positive charge accounts for the rare large-angle scattering, while the empty space explains the lack of deflection for most particles. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Rutherford’s Alpha-Particle Scattering Experiment, Impact Parameter

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion states that the impact parameter ($b$) is inversely proportional to $\sqrt{E}$. However, from the given equation, $b \propto \frac{1}{E}$ (not $\frac{1}{\sqrt{E}}$). Hence, the Assertion is false.
The Reason correctly describes the relationship between $b$, $\theta$, $Z$, and $E$ as derived from Rutherford's scattering formula. Thus, the Reason is true.
Therefore, Assertion is false, but Reason is true.

Key Concept: Scattering Angle and Impact Parameter

d) The scattering angle increases but does not double.
[Solution Description]
The relationship between the impact parameter (\$b\$) and the scattering angle (\$\theta\$) is given by: $b = \frac{1}{4\pi \epsilon_0} \cdot \frac{Ze^2 \cot(\theta/2)}{E_k}$
If \$b\$ is halved, then \$\cot(\theta/2)\$ must also be halved to maintain the equation. Therefore, \$\cot(\theta/2)\$ becomes \$\frac{1}{2}\cot(\theta_0/2)\$, where \$\theta_0\$ is the original angle.
Since \$\cot^{-1}\$ is a decreasing function, reducing \$\cot(\theta/2)\$ means \$\theta/2\$ increases, leading to a larger \$\theta\$. Thus, the scattering angle increases.

Key Concept: Number of emission transitions, Energy levels

c) 6
[Solution Description]
The number of possible emission transitions from energy level $n$ is given by $\frac{n(n-1)}{2}$.
For $n = 4$, $\frac{4 \times 3}{2} = 6$.
Hence, there are 6 possible transitions.

Key Concept: Excitation potential, photon emission

b) $255$ nm
[Solution Description]
The energy difference between the first excited state and the ground state is given as $4.86$ eV.
Using $E = \frac{hc}{\lambda}$, we can find the wavelength $\lambda$:
$\lambda = \frac{hc}{E}$.
Substituting $h = 6.626 \times 10^{-34} \, \text{J s}$, $c = 3 \times 10^8 \, \text{m/s}$, and $E = 4.86 \times 1.6 \times 10^{-19} \, \text{J}$:
$\lambda = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{4.86 \times 1.6 \times 10^{-19}}$,
$\lambda \approx 2.55 \times 10^{-7} \, \text{m} = 255 \, \text{nm}$.

Key Concept: Neutron Number Calculation

b) 6
[Solution Description]
The neutron number is given by $\text{Neutron Number} = A - Z$. Rearranging to find $Z$, we get $Z = A - \text{Neutron Number}$. Substituting $A = 12$ and $\text{Neutron Number} = 6$, $Z = 12 - 6 = 6$.

Key Concept: Atomic Stability

c) Continuous radiation causes energy loss, leading to collapse
[Solution Description]
Classical electromagnetic theory predicts that an electron orbiting the nucleus should continuously emit radiation, losing energy and spiraling inward, leading to atomic collapse within approximately $10^{-8}$ s. This contradicts observed atomic stability.

Key Concept: Alpha-Particle Scattering

b) The model predicted negligible deflection due to uniform positive charge distribution
[Solution Description]
In Thomson's model, the positive charge was spread out, so $\alpha$-particles would not experience strong repulsion and would pass through with minimal deflection. However, Rutherford observed some $\alpha$-particles deflected at large angles, indicating a concentrated positive charge. Thus, Thomson's model could not explain this phenomenon.

Key Concept: Scattering Dependence

c) The nucleus caused a single strong repulsion, explaining the $\sin^4(\theta/2)$ dependence.
[Solution Description]
Rutherford's model assumed a concentrated nucleus, where $\alpha$-particles experienced a strong Coulomb repulsion proportional to $1/r^2$ during a single close encounter. The scattering angle $\theta$ followed an inverse relationship with $\sin^4(\theta/2)$, consistent with a single high-magnitude force.
In contrast, Thomson's model predicted multiple small deflections due to distributed positive charge, leading to a different angular dependence that did not match the observed $\sin^4(\theta/2)$ trend.

Key Concept: Nuclear Structure

c) In the protons and neutrons within the nucleus
[Solution Description]
Rutherford's model states that almost the entire mass of the atom is concentrated in a tiny, dense, positively charged nucleus. The radius of the nucleus is about $10^{-15}$ m.

Key Concept: Distance of Closest Approach

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the distance of closest approach $r_0$ is inversely proportional to the initial kinetic energy $K$. This is supported by the equation $r_0 = \frac{1}{4 \pi \epsilon_0} \cdot \frac{2Ze^2}{K}$, where all other terms are constants. Thus, as $K$ increases, $r_0$ decreases, confirming the inverse proportionality.
The reason states that at the distance of closest approach, the electrostatic potential energy equals the initial kinetic energy. This is correct because the $\alpha$-particle comes to rest momentarily when all its kinetic energy is converted into electrostatic potential energy, which occurs at $r_0$.
Therefore, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: De Broglie Wavelength, Bohr's Quantization

b) 1.58 \times 10^{-10} \, \text{m}
[Solution Description]
Step 1: Calculate the momentum of the electron.
$p = m_ev = (9.11 \times 10^{-31})(2.2 \times 10^6) = 2.0042 \times 10^{-24} \, \text{kg m/s}$.
Step 2: Find the de Broglie wavelength $\lambda$.
$\lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{2.0042 \times 10^{-24}} = 3.305 \times 10^{-10} \, \text{m}$.
Step 3: For the third orbit, the circumference must be $3\lambda$ for a standing wave.
$2\pi r = 3\lambda \Rightarrow r = \frac{3 \times 3.305 \times 10^{-10}}{2\pi} \approx 1.58 \times 10^{-10} \, \text{m}$.

Key Concept: Energy levels, Emission spectrum

b) The photon has the highest energy among all possible transitions.
[Solution Description]
When an electron jumps from $n=4$ to $n=1$, the energy difference is:
$\Delta E = 13.6 \left( \frac{1}{1^2} - \frac{1}{4^2} \right) = 13.6 \left( 1 - \frac{1}{16} \right) = 13.6 \left( \frac{15}{16} \right) = 12.75 \text{ eV}$
The wavelength of the emitted photon is:
$\lambda = \frac{12375}{12.75} \approx 970 \text{ Å}, \text{ which lies in the ultraviolet region.}$
Hence, the emitted photon has the highest energy among all possible transitions in the hydrogen atom.

Key Concept: Atomic Mass Number, Neutron Number

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that the neutron number $N$ for Lithium ($^7_3 Li$) is calculated as $N = A - Z = 7 - 3 = 4$, which is correct. The reason explains that in the notation $^A_Z X$, $A$ represents the mass number (total nucleons) and $Z$ represents the atomic number (number of protons), which is also correct. However, the reason does not directly explain why the neutron number formula works; it merely defines the terms. Therefore, both are true, but the reason is not the correct explanation of the assertion.

Key Concept: Coulomb's inverse-square law verification

c) Coulomb's inverse-square law at atomic distances
[Solution Description]
The $\alpha$-particle scattering experiment verified Coulomb's inverse-square law at atomic distances. Rutherford showed that the repulsive force between $\alpha$-particles and the nucleus follows $F \propto \frac{1}{r^2}$. The number of scattered $\alpha$-particles ($N$) was found to vary with the scattering angle ($\theta$) as $N \propto \frac{1}{\sin^4(\theta/2)}$, confirming the inverse-square relationship.

Key Concept: Rutherford's Alpha-Particle Scattering Experiment

d) The atom contains a small, dense, positively charged nucleus.
[Solution Description]
Rutherford observed that most alpha-particles passed straight through the gold foil, but a few were deflected at angles greater than 90°. This indicated the presence of a small, dense, positively charged nucleus within the atom that repelled the positively charged alpha-particles strongly.

Key Concept: Radius of Bohr's Orbits

b) 2.12 \, \text{\AA}
[Solution Description]
The radius ($r_n$) of the nth Bohr orbit is given by:
$r_n = r_1 \times n^2$
where $r_1 = 0.53 \, \text{\AA}$. For $n = 2$:
$r_2 = 0.53 \times 4 = 2.12 \, \text{\AA}$

Key Concept: Nuclear size estimation, Kinetic energy

c) Nucleus appears the same size
[Solution Description]
The distance of closest approach $r_0$ gives an estimate of the nuclear size. Using:
$r_0 = \frac{1}{4\pi \epsilon_0} \cdot \frac{2Ze^2}{K}$,
if $K$ is doubled, $r_0$ becomes half. However, the actual nuclear size remains the same; only the measured $r_0$ changes. Thus, the experiment would suggest a smaller minimum distance but does not imply the nucleus itself has changed in size.

Key Concept: Total energy, Kinetic and potential energy

b) $-1$
[Solution Description]
The total energy of an electron in the $n^{th}$ orbit is:
$E_n = -\frac{13.6}{n^2} \, \text{eV}$
For $n=2$, the total energy is:
$E_2 = -\frac{13.6}{4} = -3.4 \, \text{eV}$
The kinetic energy ($K$) of the electron is equal in magnitude to the total energy but positive:
$K = +3.4 \, \text{eV}$
Thus, the ratio of kinetic energy to total energy is:
$\frac{K}{E} = \frac{3.4}{-3.4} = -1$

Key Concept: Bohr Theory Limitation

c) Incorrectly calculates ionization energies for one-electron systems.
[Solution Description]
The syllabus mentions that Bohr's theory fails to explain multi-electron atoms, fine structure of spectral lines, relative intensities of spectral lines, Zeeman/Stark effects, distribution of electrons in orbits, and wave nature of electrons. It does not mention any issue with calculating ionization energies for one-electron systems. Thus, this is not a limitation.

Key Concept: Thomson's model, Charge distribution

c) Electrons would shift slightly but stay within the positive sphere.
[Solution Description]
In Thomson's plum pudding model, electrons are embedded in a uniform positive charge sphere. When an external electric field is applied, the electrons would shift slightly from their equilibrium positions due to the force $F = eE$, but would remain bound within the positively charged sphere. This would create a temporary dipole moment until the field is removed.

Key Concept: Conclusions from Rutherford's Experiment

c) In the nucleus at the center of the atom
[Solution Description]
Rutherford concluded that the positive charge in an atom is concentrated in a very small region called the nucleus, which has a radius of about $10^{-15} \, \text{m}$.

Key Concept: Rydberg formula for hydrogen-like atoms

d) 1215 \AA
[Solution Description]
The wavelength $\lambda$ for a hydrogen-like ion is given by:
$\frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$
Substituting $Z=2$, $n_f=2$, $n_i=4$, and $R = 1.097 \times 10^7 \, \text{m}^{-1}$,
$\frac{1}{\lambda} = 1.097 \times 10^7 \times 4 \left( \frac{1}{4} - \frac{1}{16} \right) = 1.097 \times 10^7 \times 4 \times \frac{3}{16} = 8.2275 \times 10^6 \, \text{m}^{-1}$
$\lambda = \frac{1}{8.2275 \times 10^6} \approx 1.215 \times 10^{-7} \, \text{m} = 1215 \, \text{\AA}$

Key Concept: Alpha-Particle Scattering Experiment, Distance of Closest Approach

d) Assertion is false, but Reason is true.
[Solution Description]
The distance of closest approach ($r_0$) is given by:
$r_0 = \frac{1}{4\pi \epsilon_0} \cdot \frac{Ze \cdot 2e}{K}$
where $K$ is the kinetic energy of the $\alpha$-particle and $Z$ is the atomic number of the nucleus. From this equation, it is clear that $r_0 \propto \frac{1}{K}$, meaning $r_0$ is inversely proportional to the kinetic energy. Therefore, Assertion (A) is false. However, Reason (R) correctly describes the physical principle behind the derivation of $r_0$, which is based on the conservation of energy where the initial kinetic energy is converted to electrostatic potential energy at the closest approach. Thus, Reason (R) is true.

Key Concept: Thomson’s Plum-Pudding Model

b) Spread uniformly throughout the entire atom
[Solution Description]
Thomson's plum-pudding model assumes that the positive charge in an atom is uniformly distributed throughout a sphere, much like plums embedded in a pudding. The negatively charged electrons are scattered within this positively charged sphere.