Q: Q1. 1. (a) Two nuclei have mass numbers in the ratio 1:3. What is the ratio of their nuclear densities?\n(b) Two nuclei have mass numbers in the ratio 1:3. What is the ratio of their n

**(a) Ratio of nuclear densities:**\nThe nuclear density is approximately constant for all nuclei, irrespective of their mass number. This is because the volume of a nucleus is directly proportional to its mass number (V ∝ A), and the mass of a nucleus is also approximately proportional to its mass number (M ∝ A). Therefore, density (ρ = M/V) remains nearly constant.\nThus, the ratio of their nuclear densities is **1:1**.\n\n**(b) Ratio of nuclear radii:**\nThe nuclear radius (R) is related to the mass number (A) by the empirical formula: R = R₀ A^(1/3), where R₀ is an empirical constant (approximately 1.2 to 1.5 fm).\nGiven the ratio of mass numbers A₁:A₂ = 1:3.\nSo, A₁ = k and A₂ = 3k for some constant k.\n\nRatio of their nuclear radii:\nR₁ / R₂ = (R₀ A₁^(1/3)) / (R₀ A₂^(1/3))\nR₁ / R₂ = (A₁ / A₂)^(1/3)\nR₁ / R₂ = (1 / 3)^(1/3)\nR₁ / R₂ = 1 / (3^(1/3))\n\nTherefore, the ratio of their nuclear radii is **1 : (3)^(1/3)** or **1 : 1.442** (approximately).

Question 1

Q1. 1. (a) Two nuclei have mass numbers in the ratio 1:3. What is the ratio of their nuclear densities?
(b) Two nuclei have mass numbers in the ratio 1:3. What is the ratio of their n

Accepted Answer

**(a) Ratio of nuclear densities:**
The nuclear density is approximately constant for all nuclei, irrespective of their mass number. This is because the volume of a nucleus is directly proportional to its mass number (V ∝ A), and the mass of a nucleus is also approximately proportional to its mass number (M ∝ A). Therefore, density (ρ = M/V) remains nearly constant.
Thus, the ratio of their nuclear densities is **1:1**.

**(b) Ratio of nuclear radii:**
The nuclear radius (R) is related to the mass number (A) by the empirical formula: R = R₀ A^(1/3), where R₀ is an empirical constant (approximately 1.2 to 1.5 fm).
Given the ratio of mass numbers A₁:A₂ = 1:3.
So, A₁ = k and A₂ = 3k for some constant k.

Ratio of their nuclear radii:
R₁ / R₂ = (R₀ A₁^(1/3)) / (R₀ A₂^(1/3))
R₁ / R₂ = (A₁ / A₂)^(1/3)
R₁ / R₂ = (1 / 3)^(1/3)
R₁ / R₂ = 1 / (3^(1/3))

Therefore, the ratio of their nuclear radii is **1 : (3)^(1/3)** or **1 : 1.442** (approximately).

Question 2

Q2. 2. The short range attractive nuclear force acting between two nucleons would be
(a) independent of the spin of the nucleons.
(b) independent of the charge of the nucleons.
(c)

Accepted Answer

The nuclear force, also known as the strong nuclear force, is the strongest of the four fundamental forces and holds the nucleons (protons and neutrons) together in the nucleus. Its properties are crucial for understanding nuclear stability.

Let's analyze the given options based on the properties of the nuclear force:

*   **(a) independent of the spin of the nucleons.** This statement is **incorrect**. The nuclear force is known to be spin-dependent. It is stronger when the spins of the interacting nucleons are parallel and weaker when they are anti-parallel.
*   **(b) independent of the charge of the nucleons.** This statement is **correct**. The nuclear force acts equally between a proton-proton pair, a neutron-neutron pair, and a proton-neutron pair. This property is called charge independence of the nuclear force. It does not distinguish between protons and neutrons based on their charge.
*   **(c) dependent on the spin of the nucleons.** This statement is **correct**. As mentioned in (a), the strength of the nuclear force depends on the relative orientation of the spins of the interacting nucleons.
*   **(d) dependent on the charge of the nucleons.** This statement is **incorrect**. As mentioned in (b), the nuclear force is charge-independent.

Therefore, the correct statements are **(b) and (c)**.

Question 3

Q3. 3. The short range attractive nuclear force acting between two nucleons would be
(a) independent of the spin of the nucleons.
(b) independent of the charge of the nucleons.
(c)

Accepted Answer

This question is identical to Question 2. The properties of the nuclear force are fundamental to understanding nuclear physics.

Let's reiterate the properties of the nuclear force:

*   **Charge Independence:** The nuclear force acts equally between any pair of nucleons (proton-proton, neutron-neutron, proton-neutron). It does not depend on the electric charge of the nucleons. Therefore, statement (b) is correct, and statement (d) is incorrect.
*   **Spin Dependence:** The strength of the nuclear force depends on the relative orientation of the spins of the interacting nucleons. It is stronger for parallel spins and weaker for anti-parallel spins. Therefore, statement (c) is correct, and statement (a) is incorrect.

Based on these properties, the correct statements are **(b) independent of the charge of the nucleons** and **(c) dependent on the spin of the nucleons**.

Question 4

Q4. 4. Obtain the binding energy (in MeV) of a nitrogen nucleus ($_{7}^{14}N$) from the following data:
$m_H (mass \ of \ hydrogen \ atom) = 1.007825 \ u$
$m_n (mass \ of \ neutron)

Accepted Answer

To obtain the binding energy of a nitrogen nucleus ($_{7}^{14}N$), we first need to calculate the mass defect. The mass defect is the difference between the sum of the masses of its constituent nucleons (protons and neutrons) and the actual mass of the nucleus.

For the nucleus $_{7}^{14}N$:
*   Number of protons (Z) = 7
*   Number of neutrons (N) = A - Z = 14 - 7 = 7

Given data:
*   Mass of hydrogen atom ($m_H$) = 1.007825 u (Note: The mass of a hydrogen atom includes one proton and one electron. When calculating mass defect for a nucleus, we use the mass of a hydrogen atom for protons, as the electron masses cancel out when comparing with the atomic mass of the nucleus.)
*   Mass of neutron ($m_n$) = 1.008665 u
*   Mass of $_{7}^{14}N$ atom ($m(_{7}^{14}N)$) = 14.00307 u
*   Energy equivalent of 1 u = 931.5 MeV/c²

**Step 1: Calculate the expected mass of the constituent nucleons.**
Expected mass = Z × $m_H$ + N × $m_n$
Expected mass = 7 × 1.007825 u + 7 × 1.008665 u
Expected mass = 7.054775 u + 7.060655 u
Expected mass = 14.115430 u

**Step 2: Calculate the mass defect (Δm).**
Mass defect (Δm) = Expected mass - Actual mass of $_{7}^{14}N$ atom
Δm = 14.115430 u - 14.00307 u
Δm = 0.11236 u

**Step 3: Calculate the binding energy (BE).**
Binding energy (BE) = Δm × (Energy equivalent of 1 u)
BE = 0.11236 u × 931.5 MeV/u
BE = 104.64054 MeV

Rounding to a reasonable number of significant figures (based on the input data, 5-6 significant figures are appropriate):
BE ≈ **104.64 MeV**

Thus, the binding energy of the nitrogen nucleus ($_{7}^{14}N$) is approximately 104.64 MeV.

Question 5

Q5. A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin

Accepted Answer

The total nuclear energy required to separate all the neutrons and protons from each other in a 3.0 g coin made of $^ {63}_{29}Cu$ is approximately 2.536 × 10¹² J or 1.583 × 10²⁵ MeV.

Question 6

Q6. Write the nuclear reaction for $\alpha$-decay of $^ {238}_{92}U$.

Accepted Answer

The nuclear reaction for the $\alpha$-decay of $^ {238}_{92}U$ is:
$^ {238}_{92}U ightarrow ^ {234}_{90}Th + ^4_2He$

Question 7

Q7. A radioactive isotope has a half-life of T years. How long will it take for the activity to reduce to (a) 3.125%, (b) 1% of its original value?

Accepted Answer

(a) It will take 5T years for the activity to reduce to 3.125% of its original value.
(b) It will take approximately 6.64 T years for the activity to reduce to 1% of its original value.

Question 8

Q8. The half-life of $^ {238}_{92}U$ undergoing $\alpha$-decay is 4.5 × 10$^9$ years. What is the activity of 1g sample of $^ {238}_{92}U$?

Accepted Answer

The activity of a 1g sample of $^ {238}_{92}U$ is approximately 1.23 × 10⁴ Bq.

Question 9

Q9. A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin

Accepted Answer

The total nuclear energy required to separate all the neutrons and protons from each other in a 3.0 g coin made of $^ {63}_{29}Cu$ is approximately 2.68 x 10^12 J or 1.67 x 10^25 MeV.

Question 10

Q10. A radioactive isotope has a half-life of T years. How much of the isotope is left after (a) T/2 years, (b) T years, (c) 2T years, and (d) 10T years?

Accepted Answer

Let $N_0$ be the initial amount of the isotope. The amount remaining after time 't' is given by $N = N_0 (1/2)^{t/T}$, where T is the half-life.

(a) After T/2 years: $N = N_0 (1/2)^{(T/2)/T} = N_0 (1/2)^{1/2} = N_0 / \sqrt{2} \approx 0.707 N_0$
(b) After T years: $N = N_0 (1/2)^{T/T} = N_0 (1/2)^1 = N_0 / 2$
(c) After 2T years: $N = N_0 (1/2)^{2T/T} = N_0 (1/2)^2 = N_0 / 4$
(d) After 10T years: $N = N_0 (1/2)^{10T/T} = N_0 (1/2)^{10} = N_0 / 1024$

Question 11

Q11. The half-life of $^ {238}U$ undergoing $\alpha$-decay is 4.5 x 10$^9$ years. What is the activity of 1 g sample of $^ {238}U$?

Accepted Answer

The activity of a 1 g sample of $^ {238}U$ is approximately 1.23 x 10$^4$ Bq or 0.33 mCi.

Question 12

Q12. Sodium-24 ($^ {24}_{11}Na$) has a half-life of 15 h. A sample of sodium-24 with an initial activity of 10 mCi is injected into a patient. What is the activity after 30 h?

Accepted Answer

The activity of the sodium-24 sample after 30 hours will be 2.5 mCi.

Question 13

Q13. A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin

Accepted Answer

To calculate the nuclear energy required to separate all neutrons and protons, we need to determine the total binding energy of all the nuclei in the coin. This involves calculating the mass defect for a single $^ {63}_{29}Cu$ nucleus, converting it to binding energy, and then multiplying by the total number of nuclei in the 3.0 g coin.

Question 14

Q14. A radioactive isotope has a half-life of T years. How much of the initial activity will remain after (a) T/2 years? (b) 2T years?

Accepted Answer

The activity of a radioactive sample decreases exponentially with time. The fraction of initial activity remaining after a certain time 't' can be calculated using the formula derived from the radioactive decay law.

Question 15

Q15. The half-life of $^ {198}_{79}Au$ is 2.7 days. Calculate (a) the decay constant, (b) the average life, and (c) the activity of 1.0 mg of $^ {198}_{79}Au$. Take atomic mass of $^ {1

Accepted Answer

This question involves calculating key parameters of radioactive decay: decay constant, average life, and activity, given the half-life and mass of the sample. We will use the relationships between these quantities and Avogadro's number.

Question 16

Q16. A radioactive sample has a half-life of 15 years. What fraction of the sample will remain undecayed after 45 years?

Accepted Answer

To find the fraction of the sample remaining undecayed, we need to determine how many half-lives have passed during the given time period and then apply the radioactive decay law.

Question 17

Q17. A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin

Accepted Answer

The nuclear energy required to separate all the neutrons and protons from each other is equivalent to the total binding energy of all the nuclei in the coin. This is calculated by first finding the mass defect for a single $^ {63}_{29}Cu$ nucleus, then converting it to binding energy per nucleus, and finally multiplying by the total number of nuclei in the coin.

Given:
Mass of coin, M = 3.0 g
Atomic mass of $^ {63}_{29}Cu$, $m_{Cu}$ = 62.92960 u
Mass of proton, $m_p$ = 1.007825 u
Mass of neutron, $m_n$ = 1.008665 u

For $^ {63}_{29}Cu$:
Number of protons, Z = 29
Number of neutrons, N = A - Z = 63 - 29 = 34

1.  **Calculate the theoretical mass of the nucleons:**
    Mass of 29 protons = $29 	imes 1.007825 	ext{ u} = 29.226925 	ext{ u}$
    Mass of 34 neutrons = $34 	imes 1.008665 	ext{ u} = 34.294610 	ext{ u}$
    Total mass of nucleons = $29.226925 	ext{ u} + 34.294610 	ext{ u} = 63.521535 	ext{ u}$

2.  **Calculate the mass defect ($\Delta m$) for one $^ {63}_{29}Cu$ nucleus:**
    $\Delta m = (	ext{Total mass of nucleons}) - (	ext{Atomic mass of } ^ {63}_{29}Cu)$
    $\Delta m = 63.521535 	ext{ u} - 62.92960 	ext{ u} = 0.591935 	ext{ u}$

3.  **Calculate the binding energy (BE) per nucleus:**
    Using the conversion factor 1 u = 931.5 MeV/c$^2$ (or 931.5 MeV for energy):
    BE per nucleus = $\Delta m 	imes 931.5 	ext{ MeV/u}$
    BE per nucleus = $0.591935 	ext{ u} 	imes 931.5 	ext{ MeV/u} = 551.39 	ext{ MeV}$

4.  **Convert BE per nucleus to Joules:**
    1 MeV = $1.602 	imes 10^{-13} 	ext{ J}$
    BE per nucleus = $551.39 	imes 1.602 	imes 10^{-13} 	ext{ J} = 8.833 	imes 10^{-11} 	ext{ J}$

5.  **Calculate the number of atoms in the coin:**
    Molar mass of Cu-63 is approximately 63 g/mol.
    Number of moles in 3.0 g = $\frac{	ext{Mass}}{	ext{Molar Mass}} = \frac{3.0 	ext{ g}}{63 	ext{ g/mol}} = 0.04762 	ext{ mol}$
    Number of atoms (N) = Number of moles $	imes$ Avogadro's number ($N_A$)
    $N = 0.04762 	ext{ mol} 	imes 6.022 	imes 10^{23} 	ext{ atoms/mol} = 2.868 	imes 10^{22} 	ext{ atoms}$

6.  **Calculate the total nuclear energy required:**
    Total energy = Number of atoms $	imes$ BE per nucleus
    Total energy = $2.868 	imes 10^{22} 	imes 8.833 	imes 10^{-11} 	ext{ J}$
    Total energy = $2.533 	imes 10^{12} 	ext{ J}$

Therefore, the nuclear energy required to separate all the neutrons and protons from each other in the 3.0 g coin is approximately $2.53 	imes 10^{12} 	ext{ J}$.

Question 18

Q18. A radioactive isotope has a half-life of T years. How much of the isotope is left after (a) T years, (b) 2T years, and (c) 9T years?

Accepted Answer

Let $N_0$ be the initial amount (or number of nuclei) of the radioactive isotope.
The half-life ($T_{1/2}$) is given as T years.

After one half-life, the amount of the isotope remaining is half of the initial amount. After 'n' half-lives, the amount remaining is given by the formula:
$N = N_0 \left(\frac{1}{2}ight)^n$
where $n = \frac{	ext{Total time (t)}}{	ext{Half-life (T)}}$

(a) **After T years:**
Here, total time $t = T$ years.
Number of half-lives, $n = \frac{T}{T} = 1$.
Amount left, $N = N_0 \left(\frac{1}{2}ight)^1 = \frac{N_0}{2}$.
So, **1/2 of the initial amount** of the isotope is left after T years.

(b) **After 2T years:**
Here, total time $t = 2T$ years.
Number of half-lives, $n = \frac{2T}{T} = 2$.
Amount left, $N = N_0 \left(\frac{1}{2}ight)^2 = N_0 	imes \frac{1}{4} = \frac{N_0}{4}$.
So, **1/4 of the initial amount** of the isotope is left after 2T years.

(c) **After 9T years:**
Here, total time $t = 9T$ years.
Number of half-lives, $n = \frac{9T}{T} = 9$.
Amount left, $N = N_0 \left(\frac{1}{2}ight)^9 = N_0 	imes \frac{1}{512} = \frac{N_0}{512}$.
So, **1/512 of the initial amount** of the isotope is left after 9T years.

Question 19

Q19. The half-life of $^ {198}_{79}Au$ is 2.7 days. Calculate (a) the decay constant, (b) the average life, and (c) the activity of 1.0 mg of $^ {198}_{79}Au$.

Accepted Answer

Given:
Half-life ($T_{1/2}$) of $^ {198}_{79}Au$ = 2.7 days
Mass of $^ {198}_{79}Au$ = 1.0 mg = $1.0 	imes 10^{-3}$ g
Molar mass of $^ {198}_{79}Au$ is approximately 198 g/mol.
Avogadro's number ($N_A$) = $6.022 	imes 10^{23}$ atoms/mol

(a) **Decay constant ($\lambda$):**
The relationship between half-life and decay constant is:
$T_{1/2} = \frac{\ln 2}{\lambda}$
So, $\lambda = \frac{\ln 2}{T_{1/2}}$
$\ln 2 \approx 0.693$

First, convert half-life to seconds for standard units:
$T_{1/2} = 2.7 	ext{ days} = 2.7 	imes 24 	ext{ hours/day} 	imes 60 	ext{ minutes/hour} 	imes 60 	ext{ seconds/minute}$
$T_{1/2} = 2.7 	imes 86400 	ext{ s} = 233280 	ext{ s}$

$\lambda = \frac{0.693}{233280 	ext{ s}} = 2.970 	imes 10^{-6} 	ext{ s}^{-1}$

(b) **Average life ($	au$):**
The average life is the reciprocal of the decay constant:
$	au = \frac{1}{\lambda}$
$	au = \frac{1}{2.970 	imes 10^{-6} 	ext{ s}^{-1}} = 336700 	ext{ s}$

To express in days:
$	au = \frac{336700 	ext{ s}}{86400 	ext{ s/day}} = 3.90 	ext{ days}$

(c) **Activity (A) of 1.0 mg of $^ {198}_{79}Au$:**
Activity is given by $A = \lambda N$, where N is the number of radioactive nuclei.

First, calculate the number of nuclei (N) in 1.0 mg of $^ {198}_{79}Au$:
Number of moles = $\frac{	ext{Mass}}{	ext{Molar Mass}} = \frac{1.0 	imes 10^{-3} 	ext{ g}}{198 	ext{ g/mol}} = 5.0505 	imes 10^{-6} 	ext{ mol}$

Number of nuclei (N) = Number of moles $	imes$ Avogadro's number ($N_A$)
$N = 5.0505 	imes 10^{-6} 	ext{ mol} 	imes 6.022 	imes 10^{23} 	ext{ atoms/mol}$
$N = 3.041 	imes 10^{18} 	ext{ nuclei}$

Now, calculate the activity:
$A = \lambda N = (2.970 	imes 10^{-6} 	ext{ s}^{-1}) 	imes (3.041 	imes 10^{18})$
$A = 9.031 	imes 10^{12} 	ext{ Bq}$ (Becquerel, or disintegrations per second)

Summary of results:
(a) Decay constant ($\lambda$) = $2.97 	imes 10^{-6} 	ext{ s}^{-1}$
(b) Average life ($	au$) = $3.90 	ext{ days}$ (or $3.37 	imes 10^5 	ext{ s}$)
(c) Activity (A) = $9.03 	imes 10^{12} 	ext{ Bq}$

Question 20

Q20. A radioactive sample has a half-life of 10 years. How long will it take for the activity to reduce to (a) 25% of its initial value, (b) 12.5% of its initial value?

Accepted Answer

Given:
Half-life ($T_{1/2}$) = 10 years.

The activity (A) of a radioactive sample is directly proportional to the number of radioactive nuclei (N) present, i.e., $A = \lambda N$. Therefore, the decay formula for activity is similar to that for the number of nuclei:
$A = A_0 \left(\frac{1}{2}ight)^n$
where $A_0$ is the initial activity, A is the activity after time t, and $n = \frac{t}{T_{1/2}}$ is the number of half-lives.

(a) **Activity reduces to 25% of its initial value:**
We want to find the time 't' when $A = 0.25 A_0$.
$0.25 A_0 = A_0 \left(\frac{1}{2}ight)^n$
$0.25 = \left(\frac{1}{2}ight)^n$
$\frac{1}{4} = \left(\frac{1}{2}ight)^n$
$\left(\frac{1}{2}ight)^2 = \left(\frac{1}{2}ight)^n$
So, $n = 2$.

Since $n = \frac{t}{T_{1/2}}$:
$2 = \frac{t}{10 	ext{ years}}$
$t = 2 	imes 10 	ext{ years} = 20 	ext{ years}$

It will take **20 years** for the activity to reduce to 25% of its initial value.

(b) **Activity reduces to 12.5% of its initial value:**
We want to find the time 't' when $A = 0.125 A_0$.
$0.125 A_0 = A_0 \left(\frac{1}{2}ight)^n$
$0.125 = \left(\frac{1}{2}ight)^n$
$\frac{1}{8} = \left(\frac{1}{2}ight)^n$
$\left(\frac{1}{2}ight)^3 = \left(\frac{1}{2}ight)^n$
So, $n = 3$.

Since $n = \frac{t}{T_{1/2}}$:
$3 = \frac{t}{10 	ext{ years}}$
$t = 3 	imes 10 	ext{ years} = 30 	ext{ years}$

It will take **30 years** for the activity to reduce to 12.5% of its initial value.

Question 21

Q21. A radioactive isotope has a half-life of T years. How long will it take for the activity to reduce to (a) 3.125%, (b) 1% of its original value?

Accepted Answer

The time taken for the activity to reduce to a certain percentage of its original value can be calculated using the radioactive decay law. The activity A at time t is given by A = A₀ (1/2)^(t/T), where A₀ is the initial activity and T is the half-life.

(a) For activity to reduce to 3.125% of its original value:
   A/A₀ = 3.125/100 = 0.03125
   0.03125 = (1/2)^(t/T)
   We know that 0.03125 = 1/32 = (1/2)⁵
   So, (1/2)⁵ = (1/2)^(t/T)
   Comparing the exponents, 5 = t/T
   Therefore, t = 5T years.

(b) For activity to reduce to 1% of its original value:
   A/A₀ = 1/100 = 0.01
   0.01 = (1/2)^(t/T)
   To solve for t/T, we take the logarithm of both sides (natural log or log base 10).
   ln(0.01) = (t/T) ln(1/2)
   ln(0.01) = (t/T) (-ln 2)
   t/T = -ln(0.01) / ln(2)
   t/T = -(-4.605) / 0.693
   t/T = 4.605 / 0.693
   t/T ≈ 6.645
   Therefore, t ≈ 6.645 T years.

Question 22

Q22. The half-life of ⁹²U²³⁸ against α-decay is 4.5 × 10⁹ years. How many disintegrations per second does 1 g of ⁹²U²³⁸ sample emit?

Accepted Answer

To find the number of disintegrations per second (activity), we need to calculate the decay constant (λ) and the number of nuclei (N) in 1 g of ⁹²U²³⁸. The activity A is given by A = λN.

Given:
Half-life (T) = 4.5 × 10⁹ years
Mass of ⁹²U²³⁸ sample (m) = 1 g
Molar mass of ⁹²U²³⁸ ≈ 238 g/mol (from the mass number)
Avogadro's number (N_A) = 6.022 × 10²³ atoms/mol

**Step 1: Convert half-life to seconds.**
T = 4.5 × 10⁹ years × (365.25 days/year) × (24 hours/day) × (3600 seconds/hour)
T ≈ 4.5 × 10⁹ × 3.15576 × 10⁷ seconds/year
T ≈ 1.420 × 10¹⁷ seconds

**Step 2: Calculate the decay constant (λ).**
λ = ln(2) / T
λ = 0.693 / (1.420 × 10¹⁷ s)
λ ≈ 4.880 × 10⁻¹⁸ s⁻¹

**Step 3: Calculate the number of nuclei (N) in 1 g of ⁹²U²³⁸.**
Number of moles = mass / molar mass = 1 g / 238 g/mol ≈ 0.0042017 mol
Number of nuclei (N) = Number of moles × Avogadro's number
N = (1 g / 238 g/mol) × 6.022 × 10²³ atoms/mol
N ≈ 0.0042017 × 6.022 × 10²³
N ≈ 2.530 × 10²¹ nuclei

**Step 4: Calculate the activity (A).**
A = λN
A = (4.880 × 10⁻¹⁸ s⁻¹) × (2.530 × 10²¹ nuclei)
A ≈ 1.234 × 10⁴ disintegrations/second

Thus, 1 g of ⁹²U²³⁸ sample emits approximately 1.234 × 10⁴ disintegrations per second.

Question 23

Q23. A given sample contains two radioisotopes A and B with half-lives 𝑇_A = 1 h and 𝑇_B = 2 h respectively. Initially, the sample contains equal numbers of both nuclei. After 2 hours

Accepted Answer

Let N₀ be the initial number of nuclei for both radioisotopes A and B, as they are initially equal.

For radioisotope A:
Initial number of nuclei = N₀
Half-life (T_A) = 1 hour
Time elapsed (t) = 2 hours
Number of half-lives for A (n_A) = t / T_A = 2 h / 1 h = 2
Number of nuclei of A remaining (N_A) = N₀ (1/2)^(n_A) = N₀ (1/2)² = N₀ / 4

For radioisotope B:
Initial number of nuclei = N₀
Half-life (T_B) = 2 hours
Time elapsed (t) = 2 hours
Number of half-lives for B (n_B) = t / T_B = 2 h / 2 h = 1
Number of nuclei of B remaining (N_B) = N₀ (1/2)^(n_B) = N₀ (1/2)¹ = N₀ / 2

Now, find the ratio of the number of nuclei of A to the number of nuclei of B left:
Ratio = N_A / N_B = (N₀ / 4) / (N₀ / 2)
Ratio = (N₀ / 4) × (2 / N₀)
Ratio = 2 / 4 = 1 / 2

So, the ratio of the number of nuclei of A to the number of nuclei of B left in the sample is 1:2.

The correct option is (b) 1:2.

Question 24

Q24. The half-life of ⁹⁰Sr³⁸ is 28 years. How much of 15 mg of this isotope will remain after 84 years?

Accepted Answer

To determine the amount of ⁹⁰Sr³⁸ remaining, we use the radioactive decay law, which relates the initial mass, half-life, and elapsed time to the remaining mass.

Given:
Initial mass (m₀) = 15 mg
Half-life (T) = 28 years
Time elapsed (t) = 84 years

**Step 1: Calculate the number of half-lives (n).**
The number of half-lives (n) is given by the total time elapsed divided by the half-life:
n = t / T
n = 84 years / 28 years
n = 3

**Step 2: Apply the decay formula for remaining mass.**
The mass (m) remaining after 'n' half-lives is given by:
m = m₀ (1/2)ⁿ
m = 15 mg × (1/2)³
m = 15 mg × (1/8)
m = 15 / 8 mg
m = 1.875 mg

Therefore, 1.875 mg of ⁹⁰Sr³⁸ will remain after 84 years.

Question 25

Q25. A radioactive isotope has a half-life of T years. How long will it take for the activity to reduce to (a) 3.125%, (b) 1% of its original value?

Accepted Answer

The time taken for the activity to reduce to:
(a) 3.125% of its original value is 5T years.
(b) 1% of its original value is approximately 6.64T years.

Question 26

Q26. The half-life of 238U undergoing α-decay is 4.5 × 10^9 years. What is the activity of 1g sample of 238U?

Accepted Answer

The activity of a 1g sample of 238U is approximately 1.23 × 10^4 Bq (or disintegrations per second).

Question 27

Q27. Carbon-14 decays with a half-life of 5730 years. What is the age of an ancient piece of wood if it is found to have only 80% of the C-14 found in a living tree?

Accepted Answer

The age of the ancient piece of wood is approximately 1845 years.

Question 28

Q28. A source contains two phosphorous radioisotopes $_{15}^{32}P$ and $_{15}^{33}P$. $_{15}^{32}P$ has a half-life of 14.3 days, while $_{15}^{33}P$ has a half-life of 25.3 days. Initi

Accepted Answer

The time one must wait until 90% of the decays come from $_{15}^{33}P$ is approximately 208.5 days.

Question 29

Q29. Under certain circumstances, a nucleus can decay by emitting a particle that is more massive than an $\alpha$-particle. Consider the following decay processes:
$_{88}^{223}Ra ig

Accepted Answer

The Q-value for the first decay ($_{88}^{223}Ra ightarrow _{82}^{209}Pb + _{6}^{14}C$) is approximately 31.8 MeV.
The Q-value for the second decay ($_{88}^{223}Ra ightarrow _{86}^{219}Rn + _{2}^{4}He$) is approximately 5.98 MeV.
Since the Q-value for the first decay (31.8 MeV) is significantly higher than that for the second decay (5.98 MeV), the decay involving the emission of a $_{6}^{14}C$ particle is energetically more favourable.

Question 30

Q30. The nuclide ¹¹C decays according to ¹¹₆C → ¹¹₅B + e⁺ + ν: T₁/₂ = 20.3 min. The maximum energy of the emitted positron is 0.960 MeV. Given the mass values:
m(¹¹₆C) = 11.011434 u
m

Accepted Answer

The Q-value for the given positron (β⁺) decay is calculated to be 0.961 MeV. This value is in excellent agreement with the measured maximum kinetic energy of the emitted positron, which is 0.960 MeV. The Q-value represents the total energy released in the decay. This energy is shared among the positron, the neutrino, and the recoiling daughter nucleus. The maximum kinetic energy of the positron occurs when the neutrino carries away negligible energy. Therefore, the Q-value is approximately equal to the maximum kinetic energy of the positron. The slight difference is attributed to the recoil energy of the daughter nucleus and experimental uncertainties.

Question 31

Q31. A radioactive isotope has a half-life of T years. How much of the isotope is left after (i) T years, (ii) 2T years, (iii) 10T years, (iv) nT years?

Accepted Answer

The amount of a radioactive isotope remaining after a certain time can be calculated using the formula for radioactive decay, which is based on the concept of half-life.

Let N₀ be the initial amount of the radioactive isotope.
Let T be the half-life of the isotope.
Let t be the total time elapsed.

The number of half-lives (n) that have passed is given by n = t / T.

The amount of the isotope remaining (N) after time t is given by the formula:
N = N₀ (1/2)^n
or N = N₀ (1/2)^(t/T)

(i) After T years:
Here, t = T years.
Number of half-lives, n = t / T = T / T = 1.
Amount remaining, N = N₀ (1/2)^1 = N₀ / 2.
So, half of the initial amount is left.

(ii) After 2T years:
Here, t = 2T years.
Number of half-lives, n = t / T = 2T / T = 2.
Amount remaining, N = N₀ (1/2)^2 = N₀ (1/4) = N₀ / 4.
So, one-fourth of the initial amount is left.

(iii) After 10T years:
Here, t = 10T years.
Number of half-lives, n = t / T = 10T / T = 10.
Amount remaining, N = N₀ (1/2)^10 = N₀ / 1024.
So, 1/1024th of the initial amount is left.

(iv) After nT years:
Here, t = nT years.
Number of half-lives, n_actual = t / T = nT / T = n.
Amount remaining, N = N₀ (1/2)^n.
So, (1/2)^n times the initial amount is left.

Chapter 13: Nuclei