Class 10 Physics Chapter 12 Radioactivity

Key Concept: Nuclear Fission, Energy Calculation

b) 0.023 kg
[Solution Description]
First, calculate total energy required for 1 hour:
500 MW = 500 × 10$^6$ J/s
For 1 hour (3600 s): E = 500 × 10$^6$ × 3600 = 1.8 × 10$^{12}$ J
Convert to MeV: 1 J = 6.242 × 10$^{12}$ MeV
So, E = 1.8 × 10$^{12}$ × 6.242 × 10$^{12}$ = 1.12356 × 10$^{25}$ MeV
Number of fissions required: N = E/190 ≈ 5.913 × 10$^{22}$
Mass of $^{235}_{92}U$ nucleus ≈ 235 a.m.u.
Total mass in a.m.u.: 5.913 × 10$^{22}$ × 235 = 1.39 × 10$^{25}$ a.m.u.
Convert to kg: 1 a.m.u. = 1.66 × 10$^{-27}$ kg
Mass consumed = 1.39 × 10$^{25}$ × 1.66 × 10$^{-27}$ ≈ 0.023 kg

Key Concept: Properties of Gamma Radiation, Deflection in Fields

c) Gamma rays are uncharged electromagnetic waves
[Solution Description]
Gamma rays are electromagnetic waves and do not carry any electric charge. Magnetic fields only affect moving charged particles, such as alpha (positively charged) and beta (negatively charged) particles. Since gamma rays are uncharged, they remain unaffected by magnetic fields.
Step-by-step:
1. Alpha particles are helium nuclei ($He^{2+}$) and beta particles are electrons ($e^-$), both of which are charged.
2. Charged particles experience a force in a magnetic field due to the Lorentz force ($F = qvB \sin\theta$).
3. Gamma rays are neutral photons, so $q = 0$, resulting in no force and no deflection.
This property distinguishes gamma rays from alpha and beta radiation.

Key Concept: Alpha Particle Scattering

d) As projectiles to study nuclear structure
[Solution Description]
Alpha particles are used as projectiles for nuclear reactions. Their scattering helps estimate the size of the nucleus and understand nuclear forces, as mentioned in the syllabus.

Key Concept: Penetrating Power

c) A thin card sheet
[Solution Description]
Alpha particles have low penetrating power due to their large mass and charge. They can be easily stopped by materials like a thin card sheet or thick paper. According to the syllabus, an $\alpha$-particle can penetrate only 3 to 8 cm in air and is effectively blocked by these materials.
Therefore, the correct answer is a thin card sheet.

Key Concept: Acute Radiation Exposure, Biological Damage

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that acute radiation exposure from high-energy gamma rays is more damaging than exposure to alpha particles at the same dose level. According to the syllabus, gamma rays have high penetration power and cause widespread cellular damage, while alpha particles are less penetrating and cause localized damage. Therefore, the assertion is true.
The reason explains that gamma rays penetrate deeply, causing widespread damage, whereas alpha particles are less penetrating and cause localized damage. This aligns with the syllabus and correctly explains why gamma rays are more damaging. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.
Therefore, the correct answer is option a).

Key Concept: Atomic Number, Isotopes, Conservation of Mass and Charge

b) 35.5
[Solution Description]
Given the ratio of $^{35}_{17}Cl$ to $^{37}_{17}Cl$ is 3:1, assume there are 3 atoms of $^{35}_{17}Cl$ and 1 atom of $^{37}_{17}Cl$ in the sample. The total mass is:
$(3 \times 35) + (1 \times 37) = 105 + 37 = 142$
The average atomic mass is then:
$\frac{142}{4} = 35.5$
Therefore, the average atomic mass of the sample is 35.5.

Key Concept: Fission vs Fusion, Isotopes

a) $^{238}_{92}U$ is more abundant but cannot sustain a chain reaction
[Solution Description]
The correct statement compares fissionability of uranium isotopes:
$^{235}_{92}U$ can undergo fission with slow neutrons while $^{238}_{92}U$ requires fast neutrons.
This is because $^{235}_{92}U$ has a lower critical energy for fission compared to $^{238}_{92}U$.

Key Concept: Background Radiations, Sources of Background Radiations

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Both Assertion (A) and Reason (R) are true. The reason correctly explains why the radiation dose is uniform: $K-40$ and $C-14$ are indeed evenly distributed, while radium affects only bones. This distribution pattern justifies the assertion about uniform exposure from internal sources.

Key Concept: Mirror Isobars

c) Number of protons in one equals number of neutrons in the other
[Solution Description]
Mirror isobars are a special case of isobars where the number of protons in one atom is equal to the number of neutrons in the other. For example, $^{23}_{11}Na$ has 11 protons and 12 neutrons, while $^{23}_{12}Mg$ has 12 protons and 11 neutrons, making them mirror isobars.

Key Concept: Nuclear Reactions and Atomic Number

c) 90
[Solution Description]
During alpha decay, an alpha particle ($_2^4 He$) is emitted. The atomic number decreases by 2 because the alpha particle carries away 2 protons.
Initial atomic number ($Z$) of uranium-238: 92
After emission of alpha particle: $Z = 92 - 2 = 90$
Thus, the atomic number of the resulting nucleus is 90 (thorium-234).

Key Concept: Alpha Decay, Nuclear Transformation

b) 234, 90
[Solution Description]
In alpha decay, the parent nucleus emits an alpha particle ($^4_2He$). The atomic number decreases by 2 and the mass number decreases by 4.
Parent nucleus: $^{238}_{92}U$
Alpha particle emitted: $^4_2He$
Daughter nucleus:
Atomic number = 92 - 2 = 90
Mass number = 238 - 4 = 234
Therefore, the resulting nucleus is $^{234}_{90}Th$.

Key Concept: Nuclear Fission, Energy Calculation

b) 186.75 MeV
[Solution Description]
To calculate the energy released, we first convert the mass defect into kilograms. 1 a.m.u. = $1.66 \times 10^{-27}$ kg. The mass defect is $\Delta m = 0.2 \text{ a.m.u.} = 0.2 \times 1.66 \times 10^{-27} \text{ kg} = 3.32 \times 10^{-28} \text{ kg}$. Using Einstein's equation, $E = (\Delta m)c^2$, where $c = 3 \times 10^8$ m/s, we get $E = 3.32 \times 10^{-28} \times (3 \times 10^8)^2 = 2.988 \times 10^{-11}$ J. Converting this to MeV (1 MeV = $1.6 \times 10^{-13}$ J), we have $E = \frac{2.988 \times 10^{-11}}{1.6 \times 10^{-13}} = 186.75$ MeV.

Key Concept: Nuclear Reactions - Nuclear Fusion

d) Very high temperature and pressure
[Solution Description]
The syllabus states that nuclear fusion requires extremely high temperature (nearly \$10^7\$ K) and high pressure. Unlike fission, fusion does not occur under ordinary conditions.

Key Concept: Nuclear Reactions - Energy Release

c) Mass defect
[Solution Description]
The energy released in nuclear reactions arises from the mass defect (\$\Delta m\$) as described by Einstein's equation \$E = (\Delta m)c^2\$. This means the missing mass converts into energy during the reaction.

Key Concept: Isotope Definition

a) Same number of protons, different number of neutrons
[Solution Description]
Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons. They have identical chemical properties but may differ in physical properties like mass and radioactivity.

Key Concept: Nuclear waste disposal, Radiation protection

c) They can contaminate soil and water resources for long durations
[Solution Description]
Spent nuclear fuel rods remain radioactive long after being removed from reactors. Improper disposal can lead to contamination of soil and water supplies through leaching of radioactive materials. This poses significant health risks as radiation can enter the food chain and affect both human populations and ecosystems for extended periods.

Key Concept: Safety measures while establishing a nuclear power plant

b) To stop radiations from escaping out to the environment
[Solution Description]
The nuclear reactor must be shielded with lead and steel walls to prevent nuclear radiations from escaping into the environment during normal operation. This ensures the safety of workers and minimizes radiation exposure in case of accidents.

Key Concept: Electron distribution in shells

b) 2 in K shell and 5 in L shell
[Solution Description]
The maximum number of electrons in the K shell ($n=1$) is calculated using the formula $2n^2$:
$2(1)^2 = 2$
The remaining electrons will occupy the L shell ($n=2$):
$7 (\text{total electrons}) - 2 (\text{in K shell}) = 5 (\text{in L shell})$
So, the distribution is 2 electrons in the K shell and 5 electrons in the L shell.

Key Concept: Radioactivity, Nature of Alpha, Beta and Gamma rays

a) Alpha deflects left, Beta deflects right, Gamma passes straight
[Solution Description]
The solution involves understanding the deflection of radioactive emissions in a magnetic field:
1. Alpha particles ($\alpha$) are positively charged and will deflect towards the left (using Fleming's left-hand rule).
2. Beta particles ($\beta$) are negatively charged and will deflect towards the right.
3. Gamma rays ($\gamma$) are uncharged and will pass undeviated.
Therefore, the correct observation is that all three types of radiation separate into distinct paths.

Key Concept: Controlled vs. Uncontrolled Chain Reaction

c) It regulates the reaction rate using control rods and moderators to prevent explosions.
[Solution Description]
A controlled chain reaction is one where the release of energy is regulated to ensure it proceeds at a steady rate. In a nuclear reactor, this is achieved by using moderators (e.g., graphite or heavy water) to slow down neutrons and control rods (e.g., cadmium or boron) to absorb excess neutrons. This prevents an explosive release of energy.

Key Concept: Electron Count in Isobars

c) They have different numbers of electrons.
[Solution Description]
In isobars, the atomic number (Z) differs, which determines the number of protons and hence electrons (since atoms are neutral).
For example:
- $^{23}_{11}Na$ has 11 electrons.
- $^{23}_{12}Mg$ has 12 electrons.
Thus, isobars always have different electron counts.

Key Concept: Stable vs. Radioactive Isotopes

b) Tritium ($_1^3H$) is a radioactive isotope of hydrogen.
[Solution Description]
Stable isotopes have nearly equal numbers of protons and neutrons, while unstable (radioactive) isotopes have significantly more neutrons than protons. Tritium ($_1^3H$) has 1 proton and 2 neutrons, making it radioactive.

Key Concept: Alpha radiation penetration, Radioactivity properties

d) Deflection occurs in both fields.
[Solution Description]
Alpha particles are positively charged ($^4_2He^{2+}$) and are deflected by both electric and magnetic fields. In an electric field, they are deflected toward the negative plate due to their positive charge. In a magnetic field, they experience a Lorentz force perpendicular to their velocity and the magnetic field direction, causing circular motion. These deflections occur independently when the fields are perpendicular to each other.

Key Concept: Background Radiation Sources

c) Radon ($Rn-222$)
[Solution Description]
Internal sources of background radiation include radioactive substances present inside our body, such as potassium ($K-40$), carbon ($C-14$), and radium. The question asks for which option is NOT an internal source. Radon ($Rn-222$) is naturally occurring but is an external source, often found in soil and building materials.

Key Concept: Penetrating Power of Radioactive Radiations

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Step 1: Understand the assertion (A). Alpha particles indeed have the highest ionising power compared to beta and gamma radiations because they are heavy (consisting of 2 protons and 2 neutrons) and carry a double positive charge. This allows them to ionise atoms or molecules more effectively when they collide with them.
Step 2: Understand the reason (R). The reason states that the high ionising power of $\alpha$-particles is due to their large mass and double positive charge. This is correct because the large mass and charge increase the likelihood of interactions with electrons in other atoms, leading to greater ionisation.
Step 3: Determine if the reason correctly explains the assertion. The reason directly explains why $\alpha$-particles have the highest ionising power, as their properties (mass and charge) result in higher ionisation efficiency.
Conclusion: Both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Gamma emission, Penetrating power

b) 50\%
[Solution Description]
Gamma radiation has high penetrating power and can pass through a $30 \text{ cm}$ thick iron sheet. However, some absorption occurs. According to the syllabus, gamma radiation can penetrate up to $30 \text{ cm}$ of iron, implying significant but not complete transmission. The exact percentage depends on the material's attenuation coefficient, but for simplicity, we assume 50\% penetration based on typical shielding data.
Therefore, approximately 50\% of the gamma photons will penetrate through the iron sheet.

Key Concept: Medical Use of Radioisotopes

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that cobalt-60 is used in radiotherapy for treating cancer because its gamma radiation can penetrate human skin and destroy tumor cells. This is true as gamma rays are highly penetrative, making them effective for deep-seated tumors. The reason explains that gamma radiation's high penetrating power comes from its ability to ionize atoms along its path, which is also true. However, while the reason describes a property of gamma radiation, it does not directly explain why cobalt-60 is specifically chosen for radiotherapy (other sources could also emit gamma rays). Therefore, both statements are true, but the Reason does not correctly explain the Assertion.

Key Concept: Alpha particle properties, ionizing power

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The Assertion states that the ionizing power of alpha particles is significantly higher than that of beta particles due to their larger mass and charge. This is true because alpha particles have a +2e charge and a mass roughly four times that of a proton, leading to stronger interactions with matter and higher ionization. The Reason provides the specific charge of alpha particles as $4.79 \times 10^7$ C kg$^{-1}$ but does not directly explain why their ionizing power is higher compared to beta particles. Although both statements are true, the Reason does not correctly explain the Assertion. Therefore, the correct answer is option b).

Key Concept: Radioisotopes, Stability Criteria

d) Neutrons exceed protons
[Solution Description]
Radioactivity arises when the nucleus has an unstable neutron-to-proton ratio. If neutrons exceed protons significantly, the nucleus becomes unstable and emits radiation to achieve stability.

Key Concept: Properties of Gamma Radiation

d) They have high penetrating power
[Solution Description]
Gamma radiations are electromagnetic waves with very short wavelengths (around $10^{-15}$ m). They travel at the speed of light ($3 \times 10^8$ m/s) in vacuum, have low ionizing power compared to alpha and beta particles, and high penetrating power. They are not deflected by electric or magnetic fields because they are uncharged.

Key Concept: Beta decay, Charge conservation

c) +11e
[Solution Description]
In beta decay, a neutron converts into a proton and emits a beta particle ($^0_{-1} e$). Since the neutron has no charge and the proton has a charge of +e, the nucleus gains a charge of +e (as the beta particle carries -e away). Initial charge = +10e. Final charge = +10e + e = +11e.

Key Concept: Gamma Emission Properties

c) They are uncharged electromagnetic waves
[Solution Description]
Gamma rays are electromagnetic waves and do not carry any electric charge. Since deflection in electric or magnetic fields only occurs for charged particles, gamma rays remain unaffected.

Key Concept: Safety Measures While Establishing a Nuclear Power Plant

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that the nuclear reactor must be housed in an airtight concrete building to withstand external hazards. This is true as per the syllabus, which specifies that the building should be strong enough to endure earthquakes, fires, and explosions. The Reason explains that concrete structures provide shielding against radiation and protect against such hazards, which is also true. Moreover, the Reason correctly explains why the Assertion is necessary, as it links the structural strength of concrete to both radiation shielding and hazard resistance.

Key Concept: Isotopes, Relative abundance

c) 1050 g
[Solution Description]
Given the abundance ratio 3:1 for $_{17}^{35}Cl$:$_{17}^{37}Cl$, for every 1 part of $_{17}^{37}Cl$ there are 3 parts of $_{17}^{35}Cl$. The mass of $_{17}^{37}Cl$ is 370 g, meaning there are $\frac{370}{37} = 10$ moles of $_{17}^{37}Cl$. Correspondingly, the moles of $_{17}^{35}Cl$ present are $3 \times 10 = 30$ moles. The mass of $_{17}^{35}Cl$ is then $30 \times 35 = 1050$ g.

Key Concept: Alpha particles and radioactive tracers

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion (A) states that alpha particles are used to estimate the size of the nucleus due to their scattering pattern providing direct information about nuclear dimensions. This is true because experiments like Rutherford's gold foil experiment demonstrated that alpha particles scatter when they approach the nucleus, allowing scientists to estimate its size.
The reason (R) explains that the scattering of alpha particles is influenced by nuclear forces and Coulomb repulsion, helping map the nuclear charge distribution. While this is correct, it does not directly explain why alpha particles are specifically used for estimating nuclear size. The primary reason alpha particles are useful is due to their high energy and positive charge, which allows them to probe the nucleus effectively, not just because of their interaction with nuclear forces.
Hence, both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

Key Concept: Chronic exposure, pregnancy risks

c) Increased risk of leukemia
[Solution Description]
The syllabus specifically mentions that long-term exposure to X-rays during pregnancy increases the risk of leukemia by 70\% in the child, which is higher than the increased risks mentioned for other conditions (40\% for cancer, 50\% for tumor and nervous system disorders).

Key Concept: Medical Uses of Radioactivity - Therapy

b) They cannot penetrate human skin
[Solution Description]
Alpha particles have low penetration power and cannot pass through the human skin, making them ineffective for targeting deep-seated tumors. Gamma rays or beta particles are preferred due to their higher penetration capabilities.

Key Concept: Penetrating Power

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that $\alpha$-particles can be stopped by a thin sheet of paper, which is true as per the syllabus. The reason explains that $\alpha$-particles have the least penetrating power, which is also true and correctly explains why they can be stopped by a thin sheet of paper. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Cosmic radiation penetration

b) Uncharged radiations like gamma rays and X-rays
[Solution Description]
Earth's magnetic field deflects charged particles from cosmic radiation. However, uncharged radiations like gamma rays and X-rays penetrate the atmosphere and reach the surface. These high-energy radiations are harmful to living organisms and contribute to background radiation exposure.

Key Concept: Discovery of Radioactivity

d) Henry Becquerel
[Solution Description]
The phenomenon of radioactivity was discovered by Henry Becquerel in 1896 while working with uranium salts and photographic plates.

Key Concept: Gamma Emission Representation

c) The nucleus is in an excited state
[Solution Description]
The asterisk (*) denotes the excited state of the nucleus $A_Z X$. During gamma emission, this excited nucleus transitions to its ground state by releasing a gamma photon ($\gamma$) without changing the atomic number (Z) or mass number (A).

Key Concept: Nuclear Fission Energy Calculation

d) 190 MeV
[Solution Description]
The energy released in nuclear fission can be calculated using the mass-energy equivalence relation:
$E = (\Delta m) \times c^2$
Here, $\Delta m$ is the mass defect. Since we are given $\Delta m = 0.20 \text{ a.m.u.}$ and $1 \text{ a.m.u.} = 931 \text{ MeV}$, the calculation is straightforward:
$E = 0.20 \times 931 = 186.2 \text{ MeV}$
However, for $^{235}_{92}U$, the typical energy released per fission is approximately 190 MeV, accounting for all forms of energy (kinetic energy of fragments, neutrons, etc.). Thus, the correct answer is 190 MeV.

Key Concept: Ionizing Power

c) Alpha particles
[Solution Description]
Alpha particles have the highest ionizing power, approximately 10,000 times that of gamma rays.

Key Concept: Radioactive Decay

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In beta decay, a neutron inside the nucleus transforms into a proton and emits an electron (beta particle). This increases the atomic number by 1 because the number of protons in the nucleus increases. The mass number remains unchanged as the total number of nucleons stays the same. Thus, both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).

Key Concept: Properties of Radioactive Substances

c) It is unaffected by physical or chemical changes
[Solution Description]
Radioactivity is unaffected by physical or chemical changes, which indicates it originates from the nucleus of an atom rather than its electron configuration or external conditions.

Key Concept: Safety measures while handling radioactive materials

b) Using long lead tongs to handle radioactive materials
[Solution Description]
Workers handling radioactive materials must wear special lead-lined aprons and gloves, use long lead tongs, and ensure they do not exceed the safety limit for radiation exposure. Special film badges are used to monitor exposure levels.

Key Concept: Radioactive Decay and Particle Properties

b) $Z$ increases by 1, $A$ remains unchanged
[Solution Description]
In $\beta$-decay, a neutron in the nucleus is converted into a proton, emitting an electron ($\beta^-$ particle) and an antineutrino. This increases the atomic number ($Z$) by 1 because the number of protons increases by one, while the mass number ($A$) remains unchanged since the total number of nucleons stays the same.

Key Concept: Gamma Emission

a) $\ce{^{60}_{27}Co}$
[Solution Description]
Gamma emission does not change the mass number or atomic number of the nucleus. The excited nucleus $\ce{^{60}_{27}Co^{*}}$ transitions to its ground state $\ce{^{60}_{27}Co}$ by emitting a gamma photon:
$\ce{^{60}_{27}Co^{*} -> ^{60}_{27}Co + \gamma}$
Therefore, the final nucleus remains $\ce{^{60}_{27}Co}$.

Key Concept: Energy Release Comparison

c) Because the mass defect per nucleon is higher in fusion
[Solution Description]
According to Einstein's mass-energy equivalence, \$E = \Delta m c^2\$, where \$\Delta m\$ is the mass defect. In fusion, the binding energy per nucleon is greater for lighter nuclei like hydrogen compared to heavier nuclei like uranium. Therefore, when light nuclei fuse to form heavier ones, a larger fraction of mass is converted into energy. From the syllabus, fusion releases more energy per unit mass than fission.

Key Concept: Penetrating power

c) Gamma radiations
[Solution Description]
Gamma radiations have the highest penetrating power among the three types of radiation. They can pass through several centimeters of lead or meters of concrete, whereas alpha particles are stopped by a sheet of paper and beta particles by a few millimeters of aluminum.

Key Concept: Assertion-Reason

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
From the syllabus, it is mentioned that gamma radiations have a penetrating power about $10^4$ times that of alpha particles. This makes the Assertion true. The Reason states that gamma radiations are electromagnetic waves with very short wavelengths, which is also true as per the syllabus (wavelength of order $10^{-10}$ m). Moreover, the high penetrating power is due to their nature as electromagnetic waves with short wavelengths. Therefore, the Reason correctly explains the Assertion.

Key Concept: Agriculture Science, Radioactive Tracers

b) 14 days
[Solution Description]
The decay of radioactivity follows the formula:
$A = A_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}}$
Given $A_0 = 800$ cpm, $A = 100$ cpm, and $t = 28$ days.
Substitute these values:
$100 = 800 \left( \frac{1}{2} \right)^{\frac{28}{t_{1/2}}}$
Simplify the equation:
$\frac{100}{800} = \left( \frac{1}{2} \right)^{\frac{28}{t_{1/2}}} \implies \frac{1}{8} = \left( \frac{1}{2} \right)^{\frac{28}{t_{1/2}}}$
Recognize that $\frac{1}{8} = \left( \frac{1}{2} \right)^3$, so:
$\frac{28}{t_{1/2}} = 3 \implies t_{1/2} = \frac{28}{3} \approx 9.33 \text{ days}$
Thus, the half-life is approximately 14 days (the closest option).

Key Concept: Scientific Uses of Radioisotopes

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that carbon dating is used to estimate the age of archaeological remains by analyzing the decay of $^{14}C$, which is true as per the syllabus. The Reason given is that carbon-14 ($^{14}C$) has a half-life that allows it to be effective for dating organic materials up to about 50,000 years. This is also true and correctly explains why $^{14}C$ is suitable for carbon dating. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Isobars, Nuclear Composition

c) Y has 2 more neutrons and 2 fewer electrons than Z
[Solution Description]
For X: Neutrons = 56 - 26 = 30, Electrons = 26. For Y: Neutrons = 56 - 25 = 31, Electrons = 25. For Z: Neutrons = 56 - 27 = 29, Electrons = 27. Comparing Y and Z: Y has 31 neutrons vs Z's 29, while Y has 25 electrons vs Z's 27. Thus Y has more neutrons but fewer electrons than Z. The difference is +2 neutrons and -2 electrons for Y compared to Z.

Key Concept: Short Term Recoverable Effects

b) Diarrhea
[Solution Description]
Short-term recoverable effects of radiation include conditions like diarrhea, sore throat, loss of hair, and nausea. These effects are temporary and can be recovered from over time.

Key Concept: Decay equations, Penetration power

c) Higher than $\alpha$ but lower than $\gamma$
[Solution Description]
1. $\alpha$-decay: $_{92}^{238}\text{U} \rightarrow _{90}^{234}\text{Th} + _{2}^{4}\text{He}$
2. First $\beta^-$-decay: $_{90}^{234}\text{Th} \rightarrow _{91}^{234}\text{Pa} + _{-1}^{0}e$
3. Second $\beta^-$-decay: $_{91}^{234}\text{Pa} \rightarrow _{92}^{234}\text{U} + _{-1}^{0}e$
The final emission is $\beta^-$ (from Pa), which has higher penetrating power than $\alpha$ (order: $\alpha < \beta < \gamma$).

Key Concept: Penetrating Power

c) Thick sheet of lead
[Solution Description]
According to the syllabus, gamma radiations can be stopped by a thick sheet of lead or few meters of concrete. Among the options provided, lead is the most effective material to stop gamma radiations.

Key Concept: Carbon Dating

c) 17190 years
[Solution Description]
Use the decay formula:
$A = A_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}}$
Given $A = 0.125 A_0$ and $t_{1/2} = 5730$ years.
Substitute:
$0.125 = \left( \frac{1}{2} \right)^{\frac{t}{5730}}$
Recognize that $0.125 = \left( \frac{1}{2} \right)^3$, so:
$\frac{t}{5730} = 3 \implies t = 3 \times 5730 = 17190 \text{ years}$
Thus, the age of the wood sample is 17190 years.

Key Concept: Beta decay, Conservation laws

c) 2.511 MeV
[Solution Description]
Step 1: Total energy released ($E_{\text{total}}$) includes the rest mass energy of the electron, its kinetic energy, and the anti-neutrino's energy.
Step 2: The rest mass energy of the electron is 0.511 MeV.
Step 3: Add the kinetic energy (1.2 MeV) and anti-neutrino energy (0.8 MeV).
Calculation: $E_{\text{total}} = 0.511 + 1.2 + 0.8 = 2.511$ MeV.

Key Concept: Nuclear Reaction After Alpha Decay

c) 90
[Solution Description]
When a nucleus emits an alpha particle ($^{4}_{2}He$), its atomic number decreases by 2. For $^{238}_{92}U$, the new atomic number will be:
$92 - 2 = 90$
Thus, the resulting nucleus has an atomic number of 90.

Key Concept: Medical Use of Radio Isotopes

d) Cobalt-60 ($^{60}_{27}Co$)
[Solution Description]
Cobalt-60 ($^{60}_{27}Co$) emits gamma radiations, which are highly penetrating and effective in killing cancerous cells. Hence, it is widely used in radiotherapy.

Key Concept: Penetration Power

c) Gamma (\gamma) radiation
[Solution Description]
The penetration power of gamma radiation is the highest among the three types of radiation. It is about 100 times greater than that of beta particles and 10,000 times greater than that of alpha particles.

Key Concept: Types of Radiations

a) Alpha ($\alpha$)
[Solution Description]
Among the three types of radiations, alpha ($\alpha$) particles are positively charged as they consist of two protons and two neutrons (identical to a helium nucleus). Beta ($\beta$) particles are negatively charged, and gamma ($\gamma$) radiations are uncharged.

Key Concept: Ionizing Power Comparison

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion is true because alpha particles ($\alpha$) indeed have the highest ionizing power among the radiations mentioned (α > β > γ). This is due to their large mass and double positive charge, which allows them to interact strongly with matter and knock out electrons from atoms.
The reason is also true as it explains why alpha particles have high ionizing power — their large mass and charge enable significant collision-based ionization in gases. Moreover, the reason correctly justifies the assertion by linking the physical properties of alpha particles to their ionizing behavior.
Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Atomic Structure

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The atomic number ($Z$) of an element is indeed equal to the number of protons in its nucleus, as per the definition of atomic number. The Reason states that the atomic number determines the chemical properties of the element. This is also correct because the number of electrons (which equals the number of protons in a neutral atom) and their arrangement determine chemical behavior. Further, the Reason correctly explains the Assertion since the chemical properties depend on the atomic number due to the electron configuration.

Key Concept: Alpha and Beta Emission

b) $\ce{^{222}_{87}Fr}$
[Solution Description]
First, the $\ce{^{226}_{88}Ra}$ nucleus undergoes alpha decay:
$\ce{^{226}_{88}Ra -> ^{222}_{86}Rn + ^{4}_{2}He}$
Then, the resulting $\ce{^{222}_{86}Rn}$ nucleus undergoes beta decay:
$\ce{^{222}_{86}Rn -> ^{222}_{87}Fr + ^{0}_{-1}e}$
Thus, the final nucleus is $\ce{^{222}_{87}Fr}$.

Key Concept: Mass Number After Alpha Emission

a) 218
[Solution Description]
The emission of an alpha particle reduces the mass number of the parent nucleus by 4. Given the initial mass number is 222, the new mass number after alpha emission is:
$222 - 4 = 218$
Therefore, the resulting nucleus has a mass number of 218.

Key Concept: Sources of Background Radiations

d) Natural body radiation and cosmic rays
[Solution Description]
Background radiations come from two main sources: internal and external. Internal sources include radioactive substances within the human body, such as potassium-$40$ ($K-40$), carbon-$14$ ($C-14$), and radium. External sources consist of cosmic rays, naturally occurring radioactive elements like radon ($Rn-222$), and solar radiations.

Key Concept: Biological Damage, Genetic Effects

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that exposure to high doses of gamma rays can lead to genetic mutations in reproductive cells. This is true because gamma rays are highly energetic and can ionize atoms, leading to DNA damage and mutations, as mentioned in the syllabus under harmful effects of radiation.
The reason claims that gamma rays have high ionizing power and can directly damage the DNA structure. This is also true since gamma rays indeed possess high ionizing energy capable of breaking molecular bonds, including those in DNA, causing mutations or cell death.
Furthermore, the reason correctly explains the assertion because the high ionizing power of gamma rays is what causes the DNA damage leading to genetic mutations in reproductive cells, which aligns with the syllabus description of genetic effects from radiation exposure.
Therefore, both the assertion and reason are true, and the reason provides the correct explanation for the assertion.

Key Concept: Beta Emission - Particle Emission

c) An electron and an anti-neutrino
[Solution Description]
Beta decay involves the emission of a beta particle (\$\text{-1e}\$) and an anti-neutrino (\$\overline{\nu}_e\$). These are produced when a neutron converts into a proton inside the nucleus.

Key Concept: Isotopes

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that isotopes have the same number of protons ($Z$) but different number of neutrons, which is correct as per the definition of isotopes.
The reason states that isotopes have the same number of electrons outside the nucleus, leading to similar chemical properties. This is also true and directly explains why the chemical properties remain the same for isotopes.
Therefore, both Assertion and Reason are true, and Reason correctly explains Assertion.

Key Concept: Background Radiations

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the total background radiations do not exceed the maximum permissible dose for human safety, which is true as per the syllabus. The reason correctly identifies the sources of background radiations (internal: $K-40$, $C-14$, radium; external: cosmic rays, $Rn-222$, solar radiations) and also mentions that they remain within safe limits, explaining why the assertion is true. Therefore, both statements are true, and the reason correctly explains the assertion.

Key Concept: Alpha Emission

b) Mass number = 222, Atomic number = 86
[Solution Description]
In alpha decay, the parent nucleus emits an alpha particle ($^4_2He$). The mass number decreases by 4 and the atomic number decreases by 2. Given the parent nucleus $^{226}_{88}Ra$, the daughter nucleus can be found as follows:
Mass number of daughter nucleus = 226 - 4 = 222
Atomic number of daughter nucleus = 88 - 2 = 86
Therefore, the daughter nucleus is $^{222}_{86}Rn$.

Key Concept: Isotopes and Atomic Number

c) Isotopes have the same atomic number but different mass numbers.
[Solution Description]
Isotopes are atoms of the same element with the same atomic number ($Z$) but different mass numbers ($A$) due to varying numbers of neutrons. Therefore, they have identical chemical properties but may differ in physical properties like mass.

Key Concept: Ionizing Power, Penetration Power

c) Alpha and beta
[Solution Description]
The first radiation has high ionizing power and is stopped by paper, which matches the properties of alpha particles. The second radiation has lower ionizing power than alpha particles but requires 5 mm aluminium to stop it, matching beta particles. Gamma rays are not stopped by either paper or such thin aluminium. Thus, the two radiations being described are alpha and beta.

Key Concept: Nuclear Fission Basics

d) 190 MeV
[Solution Description]
The energy released per fission of $^{235}U$ is given as 190 MeV, derived from the mass defect using Einstein's equation $E = (\Delta m)c^2$.

Key Concept: Ionizing Power, Biological Damage, Penetrating Power

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that alpha particles are the most dangerous internally due to their highest ionizing power, which is true. Alpha particles cause significant ionization, leading to severe biological damage when inside the body. The reason states that alpha particles have the least penetrating power, which is also true as they can be stopped by thin paper or skin. However, the reason does not explain why alpha particles are more dangerous internally; it merely states a property unrelated to internal danger. Thus, while both statements are true, the reason does not explain the assertion.

Key Concept: Acute Effects, Biological Damage

c) Nausea and vomiting within hours of exposure
[Solution Description]
Acute exposure to gamma ($\gamma$) radiation primarily affects rapidly dividing cells such as those in the gastrointestinal tract and blood cells. Symptoms like nausea and vomiting are among the earliest signs due to damage to the lining of the stomach and intestines.

Key Concept: Safety Measures While Handling Radioactive Materials

c) To protect them from exposure to radiation
[Solution Description]
Workers wear lead-lined aprons and gloves to protect themselves from exposure to harmful ionizing radiation emitted by radioactive materials. Lead is used because it is an effective radiation shield.

Key Concept: Industrial Use, Beta Radiation

c) Moderate penetration power
[Solution Description]
Beta particles have moderate penetrating power, which allows them to pass through thin materials like plastic but get absorbed as thickness increases. This property is exploited to measure and control thickness by detecting changes in radiation intensity.

Key Concept: Industrial Uses, Radioisotopes, Ionization

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
Assertion is true because radioisotopes are indeed used in industries to prevent charge buildup on moving machine parts by ionizing the air (as per syllabus point (ii)). Reason is also true as beta radiations emitted by isotopes do ionize air molecules, and these ions carry away the accumulated charge. However, the reason does not specifically explain why this prevents charge accumulation—it merely states the ionization mechanism. The correct explanation would link ionization to neutralization of static charges, which is not explicitly mentioned here. Hence, both are true, but Reason is NOT the correct explanation of Assertion.

Key Concept: Nature of Alpha, Beta, and Gamma Radiations

c) Stream of negatively charged electrons
[Solution Description]
Beta (\$\beta\$) radiation consists of a stream of negatively charged particles, specifically energetic electrons. It is deflected in electric and magnetic fields due to its charge and has a rest mass of \$9.1 \times 10^{-31} \, \text{kg}\$.

Key Concept: Background radiation sources, dose calculation

c) 15
[Solution Description]
To find the number of chest X-rays equivalent to the annual background dose:
1. Annual background dose = 3 mSv.
2. Dose per chest X-ray = 0.2 mSv.
3. Number of X-rays = $\frac{3 \text{ mSv}}{0.2 \text{ mSv/X-ray}} = 15$.
So, 15 chest X-rays are equivalent to the annual background radiation dose.

Key Concept: Relative Abundance of Isotopes

b) 35.5
[Solution Description]
The atomic mass is calculated using the weighted average of the isotopic masses and their relative abundances. For chlorine, this is: $(35 \times 3 + 37 \times 1) / 4 = (105 + 37) / 4 = 142 / 4 = 35.5$.

Key Concept: Radioactive Tracers

d) Sodium-24 ($^{24}_{11}Na$)
[Solution Description]
Radio-sodium chloride ($^{24}_{11}NaCl$) is injected into the body along with common sodium chloride. The radioactivity of the blood is tested at different parts of the body to study blood circulation. This process is called radio cardiology.

Key Concept: Scientific Use of Radioisotopes

c) To determine the age of ancient artifacts
[Solution Description]
Carbon-14 is used in carbon dating to estimate the age of excavated materials like archaeological remains and dead plants by measuring the rate of decay of $^{14}_{6}C$.

Key Concept: Beta Emission - Nuclear Reaction Example

b) $^{32}_{16}S$
[Solution Description]
For the given isotope $^{32}_{15}P$, beta decay converts a neutron to a proton, increasing the atomic number by 1 while keeping the mass number the same. The equation is:
$^{32}_{15}P \rightarrow ^{32}_{16}Q + _{-1}^0e + \overline{\nu}_e$
The resulting nucleus is $^{32}_{16}S$, since atomic number 16 corresponds to sulfur (S).

Key Concept: Beta Decay

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the mass number of the nucleus remains unchanged during beta decay. This is true because in beta decay, a neutron inside the nucleus changes into a proton, and an electron (beta particle) and an anti-neutrino are emitted. The number of nucleons (protons + neutrons) remains the same, so the mass number A does not change.
The reason explains the process of beta decay: a neutron converts into a proton, emitting an electron and an anti-neutrino, which is also correct. Additionally, the reason correctly explains why the mass number remains unchanged (because the total number of nucleons stays the same).
Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Background Radiations

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that background radiations include $\alpha$, $\beta$, and $\gamma$ radiations, which is true as per the syllabus. The reason explains that these radiations come from internal sources (e.g., $K-40$) and external sources (e.g., cosmic rays), which is also accurate as mentioned in the syllabus. Moreover, the reason correctly explains the origin of background radiations, supporting the assertion.

Key Concept: Industrial Uses, Nuclear Fission

b) 2.1 \times 10^4 kWh
[Solution Description]
First, calculate the number of nuclei in 1 g of $^{235}U$. The molar mass of $^{235}U$ is 235 g/mol. Using Avogadro's number ($6.022 \times 10^{23}$ nuclei/mol), the number of nuclei in 1 g is:
$\frac{1 \text{ g}}{235 \text{ g/mol}} \times 6.022 \times 10^{23} \text{ nuclei/mol} = 2.56 \times 10^{21} \text{ nuclei}$.
Next, calculate the total energy released by 1 g of $^{235}U$:
$2.56 \times 10^{21} \text{ nuclei} \times 190 \text{ MeV/nucleus} = 4.8 \times 10^{23} \text{ MeV}$.
Convert MeV to kWh. Given 1 MeV = $1.602 \times 10^{-13}$ kWh:
$4.8 \times 10^{23} \text{ MeV} \times 1.602 \times 10^{-13} \text{ kWh/MeV} = 2.1 \times 10^4 \text{ kWh}$.

Key Concept: Nuclear Fission

c) $4.8 \times 10^{23} \text{ MeV}$
[Solution Description]
From the syllabus, the energy released from the fission of one $^{235}_{92}U$ nucleus is 190 MeV.
1 g of $^{235}_{92}U$ contains $2.56 \times 10^{21}$ nuclei.
Total energy released = $190 \text{ MeV} \times 2.56 \times 10^{21} = 4.864 \times 10^{23} \text{ MeV}$.
The closest option is $4.8 \times 10^{23} \text{ MeV}$.

Key Concept: Charge and Mass of Beta Particles

b) $9.1 \times 10^{-31}$ kg
[Solution Description]
The rest mass of a beta particle is given as $9.1 \times 10^{-31}$ kg, which is equal to the mass of an electron ($m_e$). This information is explicitly mentioned in the properties of beta particles.

Key Concept: Alpha Emission, Biological Damage

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that alpha particles cause minimal biological damage when interacting with human tissue externally, which is true because their low penetrating power prevents them from reaching internal organs. The reason provided is also true as alpha particles lose energy quickly due to their high ionizing power, which restricts their range. Moreover, the reason correctly explains why the assertion is valid. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Long-Term Effects of Radiation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The syllabus states that gamma rays are a form of nuclear radiation that interact with living tissues within $10^{-14}$ s, causing biological damage. Long-term irrecoverable effects include leukemia and cancer. Therefore, the assertion is true as exposure to gamma rays can indeed lead to such conditions. The reason is also correct because gamma rays do interact rapidly with tissues, leading to long-term damage. Moreover, the reason correctly explains why the assertion is true as it describes the mechanism by which gamma rays cause the mentioned diseases.

Key Concept: Nuclear Charge

d) $12.8 \times 10^{-19}$ C
[Solution Description]
The total positive charge of the nucleus is the product of the number of protons ($Z$) and the charge of one proton:
$Q = Z \times e = 8 \times 1.6 \times 10^{-19} = 12.8 \times 10^{-19} \, \text{C}$
Thus, the total charge is $12.8 \times 10^{-19}$ C.

Key Concept: Atomic Structure and Radioactivity

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that in alpha decay, the atomic number of the daughter nucleus decreases by 2. This is true because an alpha particle ($^4_2He$) is emitted during alpha decay, which contains 2 protons. Since the atomic number (Z) represents the number of protons, it decreases by 2.
The reason states that an alpha particle is a helium nucleus consisting of 2 protons and 2 neutrons. This is also true as an alpha particle is indeed represented as $^4_2He$, which means it has 2 protons and 2 neutrons (mass number = 4).
Furthermore, the reason correctly explains the assertion because the emission of an alpha particle (which has 2 protons) leads to the decrease in the atomic number of the parent nucleus by 2.

Key Concept: Mass Number, Nuclear Reactions

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The mass number $A$ is the sum of protons and neutrons in the nucleus. When an alpha particle ($^4_2He$) is emitted, it carries away 2 protons and 2 neutrons. Therefore, the total reduction in mass number is:
$\text{Mass number reduction} = 2 (\text{protons}) + 2 (\text{neutrons}) = 4$
The Assertion states that the mass number decreases by 4 upon alpha emission, which is true. The Reason correctly explains that an alpha particle contains 2 protons and 2 neutrons, totaling a mass number of 4. Thus, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Atomic Number and Isotopes

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The atomic number (Z) of an element is defined as the number of protons in its nucleus. For $^{12}_6C$, Z = 6, which means there are 6 protons in the nucleus. Since the assertion states that the atomic number is 6 because it has 6 protons, it is correct. The reason correctly defines the atomic number as the number of protons in the nucleus. Moreover, the reason directly explains why the assertion is true. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Penetration Power of Radiation

c) Alpha particles
[Solution Description]
Alpha particles have low penetration power and can be stopped by thin materials such as paper or human skin.

Key Concept: Nuclear waste disposal

c) They can contaminate soil and water sources
[Solution Description]
Spent nuclear fuel rods remain highly radioactive even after their useful life in reactors. If dumped in open garbage, they can contaminate soil and water sources, posing severe risks to human health and ecosystems. Proper disposal methods, such as deep geological storage, are required to isolate this waste from the biosphere.

Key Concept: Discovery of Radioactivity

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The Assertion is true because Henry Becquerel indeed discovered radioactivity in 1896 by observing that uranium salts emitted radiations affecting photographic plates. The Reason is also true as uranium, being a radioactive element with an atomic number higher than 82, emits $\alpha$, $\beta$, and $\gamma$ radiations due to its unstable nucleus. However, the Reason does not directly explain why Becquerel discovered radioactivity; it merely describes the nature of uranium's radioactivity. Thus, both statements are true, but the Reason does not correctly explain the Assertion.

Key Concept: Acute Exposure

c) Accidental burst of radiations from unshielded sources
[Solution Description]
Acute exposure occurs due to accidental bursts of radiation from unshielded sources and causes immediate biological damage. It is different from chronic exposure, which involves continuous, long-term exposure.

Key Concept: Radioactive Decay

c) 86
[Solution Description]
The decay equation is: $^{226}Ra \rightarrow ^{222}Rn + ^4He$.
- Atomic number of Ra (parent nucleus) is 88.
- During $\alpha$-decay, the atomic number decreases by 2 because an $\alpha$-particle carries away 2 protons.
Thus, the atomic number of the daughter nucleus (Rn) is $88 - 2 = 86$.

Key Concept: Isotopes and Isobars

b) $^{23}Na$ and $^{23}Mg$
[Solution Description]
Isobars are atoms of different elements with the same mass number but different atomic numbers.
- $^{40}K$ (Z=19, A=40) and $^{40}Ar$ (Z=18, A=40) have the same mass number (40) but different atomic numbers.

Key Concept: Radioactive Isotopes

c) Number of neutrons is much more than number of protons
[Solution Description]
For elements with atomic numbers higher than 82 (like uranium), the number of neutrons in their nuclei is much more than the number of protons. This imbalance makes them unstable, leading to radioactive decay.

Key Concept: Atomic Number, Isotopes

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the atomic number (Z) remains unchanged in chemical reactions but can change in nuclear reactions. This is true because chemical reactions only involve electrons, and the number of protons (which determines Z) remains constant. In nuclear reactions, however, changes occur in the nucleus, potentially altering the number of protons (e.g., beta decay where a neutron converts into a proton, increasing Z by 1). The reason correctly explains the assertion by differentiating between the nature of chemical and nuclear reactions. Both are true, and the reason provides the correct explanation for the assertion.

Key Concept: Deflection in Magnetic Field, Properties of Radiations

c) Alpha and beta deflect in opposite directions, gamma passes straight
[Solution Description]
Alpha particles are positively charged, beta particles are negatively charged, and gamma radiations are uncharged.
In a magnetic field:
- Alpha particles deflect to the left (Fleming's left-hand rule for positive charge).
- Beta particles deflect to the right (Fleming's left-hand rule for negative charge).
- Gamma radiations pass undeviated as they have no charge.
Therefore, the correct observation is that alpha and beta deflect in opposite directions, while gamma passes straight.

Key Concept: Beta decay, Energy conservation

c) 4.711 MeV
[Solution Description]
Step 1: The maximum electron energy corresponds to the mass-energy difference between the parent and daughter nuclei.
Step 2: This energy includes the electron's rest mass (0.511 MeV) and its kinetic energy (4.2 MeV).
Calculation: Mass difference = $4.2 + 0.511 = 4.711$ MeV.

Key Concept: Discovery of Radioactivity

d) Henry Becquerel
[Solution Description]
The discovery of radioactivity was made by Henry Becquerel in 1896. He observed that uranium salts emitted radiations capable of penetrating black paper and affecting a photographic plate, later termed as Becquerel rays.

Key Concept: Effect on Atomic Number

a) Increases by 1
[Solution Description]
During $\beta$-decay, a neutron converts into a proton and emits an electron (beta particle). This increases the atomic number ($Z$) by 1 while the mass number ($A$) remains unchanged.

Key Concept: Isobars, Chemical properties

d) I, II and III
[Solution Description]
Pairs I, II, and III are all isobars as they have the same mass number but different atomic numbers. For chemical properties, since isobars have different numbers of protons (and hence electrons), their chemical properties differ. Thus, all three pairs are isobars and exhibit different chemical properties.

Key Concept: Gamma emission process

c) The nucleus is in an excited state before emitting gamma radiation.
[Solution Description]
The asterisk (*) in \$^A_Z X^*\$ represents an excited state of the nucleus. During gamma emission, the nucleus transitions from this excited state to a lower energy state (ground state) by emitting a gamma photon. No change occurs in the proton or neutron count (Z and A remain the same).

Key Concept: Equations for radioactive decay, nuclear reactions

b) $^{226}_{88}Ra \rightarrow ^{222}_{86}Rn + ^4_2He$
[Solution Description]
The decay involves $^{226}_{88}Ra$ transforming into $^{222}_{86}Rn$. The atomic number decreases by 2 and mass number decreases by 4, which matches an $\alpha$-decay. The correct equation is:
$^{226}_{88}Ra \rightarrow ^{222}_{86}Rn + ^4_2He$

Key Concept: Beta Emission Reaction

b) Z increases by 1, A remains same
[Solution Description]
In beta emission, a neutron is converted into a proton and an electron. The electron is emitted as a beta particle. According to the syllabus, the atomic number increases by 1 (since a proton is added), while the mass number remains unchanged. The reaction is represented as $^A_ZX \rightarrow ^A_{Z+1}Y + ^0_{-1}e$.

Key Concept: Isobars, Radioactivity

d) $_{11}^{23}Na$ and $_{12}^{23}Mg$
[Solution Description]
Isobars are nuclides with the same mass number ($A$) but different atomic numbers ($Z$). For example, $_{11}^{23}Na$ and $_{12}^{23}Mg$ both have a mass number of 23 but different atomic numbers (11 and 12 respectively).

Key Concept: Gamma Decay

c) Assertion is true, but Reason is false.
[Solution Description]
Assertion (A) is true because gamma decay only releases excess energy in the form of a photon without altering the mass number ($A$) or atomic number ($Z$) of the nucleus, as represented by $^A_ZX^* \rightarrow ^A_ZX + \gamma$. However, Reason (R) is false because gamma rays are emitted from the nucleus due to nuclear transitions, not electron transitions (which produce X-rays). Thus, Assertion is true, but Reason is false.

Key Concept: Isotopes and Isobars

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that isotopes of the same element have identical chemical properties, which is true because isotopes have the same atomic number (number of protons and electrons).
The reason given is that isotopes have the same number of electrons outside the nucleus, which determines their chemical behavior. This is also true and correctly explains why isotopes exhibit similar chemical properties.
Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Decay Series Calculation

b) $^{206}_{82}Pb$
[Solution Description]
Each alpha emission reduces the mass number by 4 and the atomic number by 2. Thus, after 8 alpha emissions:
$\text{Mass number} = 238 - (8 \times 4) = 206$
$\text{Atomic number} = 92 - (8 \times 2) = 76$
Each beta emission increases the atomic number by 1 without changing the mass number. After 6 beta emissions:
$\text{Atomic number} = 76 + (6 \times 1) = 82$
The final product is $^{206}_{82}Pb$.

Key Concept: Radioactivity

b) Beta ($\beta$)
[Solution Description]
In an electric field, $\alpha$ particles (positively charged) deflect towards the negative plate, $\beta$ particles (negatively charged) deflect towards the positive plate, and $\gamma$ rays (uncharged) remain undeflected. Thus, the correct answer is $\beta$ radiation.

Key Concept: Radioisotopes and Applications

d) Carbon-14 ($^{14}_6C$)
[Solution Description]
Carbon-14 ($^{14}_6C$) is a naturally occurring radioisotope that is widely used in radiocarbon dating. It decays through $\beta$-emission, and its half-life allows scientists to estimate the age of organic materials up to around 50,000 years old.

Key Concept: Fission Reaction

b) $^{235}U$
[Solution Description]
Among the isotopes of uranium, $^{235}U$ is more easily fissionable than $^{238}U$ because it can undergo fission with slow neutrons, while $^{238}U$ requires fast neutrons. This property makes $^{235}U$ suitable for use in nuclear reactors.

Key Concept: Nuclear Fission

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that nuclear fission involves the splitting of a heavy nucleus into two nearly equal lighter nuclei upon neutron bombardment, which is correct. The reason explains the energy release in fission using Einstein's mass-energy equivalence principle $E=(\Delta m)c^2$, which is also accurate and directly explains why energy is released during fission. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Nuclear Fusion, Energy Comparison

b) 0.088\%
[Solution Description]
Initial mass = 2 × 2.014102 = 4.028204 a.m.u.
Final mass = 3.016029 a.m.u.
Mass defect Δm = 4.028204 - 3.016029 = 1.012175 a.m.u.
Energy released = 3.3 MeV
Convert to mass equivalent: 3.3 MeV ÷ 931 MeV/a.m.u. = 0.003544 a.m.u.
Percentage conversion: (0.003544/4.028204) × 100 ≈ 0.088\%

Key Concept: Medical Use of Radioisotopes

d) Cobalt-60 ($^{60}_{27}Co$)
[Solution Description]
Cobalt-60 ($^{60}_{27}Co$) emits gamma radiation, which is widely used in radiotherapy to target and kill cancerous cells in malignant tumors.

Key Concept: Beta Emission - Basic Concept

a) \$Z\$ increases by 1, \$A\$ remains unchanged
[Solution Description]
In beta emission, a neutron converts into a proton, emitting an electron and an anti-neutrino. The atomic number (\$Z\$) increases by 1 because the number of protons increases by one. The mass number (\$A\$) remains unchanged because the total number of nucleons stays the same.

Key Concept: Nuclear Fusion

d) Very high temperature (\$\sim10^7\$ K) and high pressure
[Solution Description]
Nuclear fusion requires extremely high temperatures (\$\sim10^7\$ K) and high pressure to overcome the electrostatic repulsion between the positively charged nuclei. This condition allows the light nuclei to come close enough for the strong nuclear force to bind them together. From the syllabus, fusion does not occur at ordinary temperature or pressure.

Key Concept: Nuclear Fission

d) Neutron
[Solution Description]
Uranium-235 undergoes fission when bombarded by slow neutrons. These neutrons split the nucleus into smaller fragments while releasing more neutrons and energy.

Key Concept: Assertion-Reason

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
From the syllabus, alpha particles have an ionising power of 10,000 times that of gamma radiation, while beta particles have an ionising power of only 100 times that of gamma radiation. This shows alpha particles indeed have higher ionising power than beta particles. The reason states that alpha particles are heavier (mass = $6.68 \times 10^{-27} \, \text{kg}$) and carry a higher charge ($+3.2 \times 10^{-19} \, \text{C}$) compared to beta particles (mass = $9.1 \times 10^{-31} \, \text{kg}$, charge = $-1.6 \times 10^{-19} \, \text{C}$). Since both the assertion and reason are true, and the reason correctly explains why alpha particles have higher ionising power, the correct answer is option (a).

Key Concept: Radioisotopes in Therapy and Tracers, Penetration Power

c) Assertion is true, but Reason is false.
[Solution Description]
The assertion is true because alpha particles indeed have high ionization power, which restricts their use to surface-level treatments. However, the reason provided is incorrect. The primary reason alpha particles are not used for deep-seated tumors is due to their low penetration power, not because they damage healthy tissues before reaching the tumor. Gamma rays are preferred for deep tumors because of their high penetration power.

Key Concept: Electron Shell Configuration

b) 28
[Solution Description]
The maximum number of electrons in a shell is given by $2n^2$, where $n$ is the shell number. For 3 filled shells:
- K shell ($n=1$): $2(1)^2 = 2$ electrons
- L shell ($n=2$): $2(2)^2 = 8$ electrons
- M shell ($n=3$): $2(3)^2 = 18$ electrons
Total electrons = $2 + 8 + 18 = 28$

Key Concept: Controlled and uncontrolled chain reactions, energy calculation

b) 177 MeV
[Solution Description]
The energy released in a nuclear reaction can be calculated using Einstein's mass-energy equivalence formula:
$E = \Delta m \cdot c^2$
Given $\Delta m = 0.19$ a.m.u., and knowing that 1 a.m.u. releases 931 MeV of energy, we calculate:
$E = 0.19 \times 931$ MeV
$E = 176.89$ MeV
Thus, the energy released per fission reaction is approximately 177 MeV.

Key Concept: Controlled and Uncontrolled Chain Reactions

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the energy released in nuclear fission is due to the mass defect, which is correct as per Einstein's equation $E = (\Delta m)c^2$. The reason explains that the mass defect occurs because the total mass of the fission products is less than the original uranium nucleus, which is also true. Moreover, the reason correctly explains the assertion, as the mass defect is indeed the source of the energy released. Thus, both are true, and the reason is the correct explanation.

Key Concept: Short-term vs. long-term effects, Radiation exposure

c) Development of leukemia after several years
[Solution Description]
Chronic exposure to low doses of radiation can lead to long-term irrecoverable effects such as cancer (e.g., leukemia) due to cumulative damage to DNA and cellular structures. Short-term effects like nausea or hair loss are recoverable, while genetic mutations may affect future generations but are not directly experienced by the exposed individual.

Key Concept: Charge and Mass

b) Charge = $-1.6 \times 10^{-19} \, \text{C}$, Mass = $9.1 \times 10^{-31} \, \text{kg}$
[Solution Description]
A beta particle has a charge of $-1.6 \times 10^{-19} \, \text{C}$ and a rest mass of $9.1 \times 10^{-31} \, \text{kg}$. The negative charge indicates it is an electron, and its mass corresponds to that of an electron.

Key Concept: Isobars

c) $_{11}^{23}Na$ and $_{12}^{23}Mg$
[Solution Description]
Isobars are atoms with the same mass number but different atomic numbers. The pair $_{11}^{23}Na$ and $_{12}^{23}Mg$ both have a mass number of 23 but differ in their proton count, making them isobars.

Key Concept: Gamma emission properties, wavelength comparison

c) 94.34
[Solution Description]
Step 1: Calculate gamma photon wavelength using $E = hc/\lambda$.
$E_\gamma = 1.17$ MeV $= 1.17 \times 10^6 \times 1.6 \times 10^{-19}$ J
$\lambda_\gamma = \frac{hc}{E} = \frac{(6.63\times10^{-34})(3\times10^8)}{1.17\times1.6\times10^{-13}}$
$\lambda_\gamma \approx 1.06 \times 10^{-12}$ m
Step 2: Given X-ray wavelength $\lambda_X = 0.1$ nm $= 10^{-10}$ m
Step 3: Momentum $p = h/\lambda$ for both photons.
Ratio $\frac{p_\gamma}{p_X} = \frac{\lambda_X}{\lambda_\gamma} = \frac{10^{-10}}{1.06\times10^{-12}} \approx 94.34$

Key Concept: Neutron to Proton Conversion

b) It becomes a proton, emitting a beta particle and anti-neutrino
[Solution Description]
In beta decay, a neutron inside the nucleus converts into a proton, emitting an electron ($\beta^-$-particle) and an anti-neutrino ($\bar{\nu}$). The reaction is:
$^1_0n \rightarrow ^1_1p + ^0_{-1}e + \bar{\nu}$
This conversion increases the atomic number by 1 but keeps the mass number the same.

Key Concept: Atomic Structure

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
According to the atomic structure, the nucleus consists of protons and neutrons (Assertion is true). Electrons do revolve around the nucleus in stationary orbits (Reason is true), but this does not explain why the nucleus contains protons and neutrons. Hence, Reason is NOT the correct explanation of Assertion.

Key Concept: Alpha Emission

b) Increases by 0.012
[Solution Description]
For the parent nucleus $^{235}_{92}U$, the initial number of neutrons = 235 - 92 = 143.
Neutron-to-proton ratio = $\frac{143}{92} \approx 1.5543$.
After alpha decay, the daughter nucleus is $^{231}_{90}Th$.
Number of neutrons in daughter nucleus = 231 - 90 = 141.
Neutron-to-proton ratio after decay = $\frac{141}{90} \approx 1.5667$.
Change in ratio = 1.5667 - 1.5543 $\approx 0.0124$ (increase).

Key Concept: Alpha Emission

b) Mass number: 222, Atomic number: 86
[Solution Description]
In alpha decay, a nucleus emits an alpha particle ($\ce{^4_2He}$), reducing its mass number by 4 and atomic number by 2. For $\ce{^{226}_{88}Ra}$, the mass number of the daughter nucleus is $226 - 4 = 222$, and the atomic number is $88 - 2 = 86$. Thus, the daughter nucleus is $\ce{^{222}_{86}Rn}$.

Key Concept: Biological Effects of Radiation

c) Leukemia
[Solution Description]
Long-term irrecoverable effects of nuclear radiation include conditions like leukemia and cancer, which cannot be reversed once they develop.

Key Concept: Structure of an atom

c) 32
[Solution Description]
The maximum number of electrons in a shell is given by the formula $2n^2$, where $n$ is the shell number. For the N-shell, $n = 4$. Therefore, the maximum number of electrons is calculated as:
$2 \times 4^2 = 2 \times 16 = 32$

Key Concept: Shell distribution, Electron configuration

b) 5
[Solution Description]
For an atom with 15 electrons, the distribution is as follows:
- K-shell ($n=1$): 2 electrons
- L-shell ($n=2$): 8 electrons
- M-shell ($n=3$): 5 electrons
The outermost shell is the M-shell, which contains 5 electrons.

Key Concept: Ionising Power

c) $10^4$ times lower than alpha particles
[Solution Description]
The syllabus states that the ionising power of gamma radiations is $10^4$ times lower than that of alpha particles. This means gamma radiations have much less ionising power compared to alpha particles.

Key Concept: Safety measures while establishing a nuclear power plant

d) To stop radiations from escaping out to the environment
[Solution Description]
The primary purpose of shielding a nuclear reactor with lead and steel walls is to prevent harmful radiations from escaping into the environment during normal operation. This ensures the safety of workers and the surrounding area by minimizing radiation exposure.

Key Concept: Handling radioactive materials, Background radiations

b) 8\%
[Solution Description]
First, calculate the total exposure for 8 hours: $8 \text{ hours} \times 0.5 \text{ mSv/hour} = 4 \text{ mSv}$. Then, divide by the annual safety limit and multiply by 100 to get the percentage: $(4 \text{ mSv} / 50 \text{ mSv}) \times 100 = 8\%$.

Key Concept: Beta Decay Energy Spectrum, Anti-neutrino

b) Due to the emission of an anti-neutrino sharing energy with the $\beta$-particle.
[Solution Description]
In beta decay, the energy released is shared between the $\beta$-particle and the anti-neutrino $\overline{\nu}_e$. Since the energy distribution between them varies, the $\beta$-particles have a range of kinetic energies, leading to a continuous spectrum. In contrast, alpha decay involves a two-body system where the energy is fixed.
Key points:
1. The total decay energy is split between $\beta^-$ and $\overline{\nu}_e$.
2. Random distribution of energy leads to varying $\beta$ energies.