1. The half life of a radioactive substance is 20 minutes. In how much time, the activity of substance drops to ${(\frac{1}{16})}^{th}$ of its initial value? ⇒ (NEET 2023)

A. 40 minutes

B. 60 minutes

C. 80 minutes

D. 20 minutes

The correct answer is option (C)

Half life $T = 20 \min$

Left fraction of activity $\frac{1}{16}$

$\begin{aligned} ∵ \frac{R}{R_{0}} & = {(\frac{1}{2})}^{t / T} \\ \frac{1}{16} & = {(\frac{1}{2})}^{t / 20} \\ {(\frac{1}{2})}^{4} & = {(\frac{1}{2})}^{t / 20} \\ 4 & = \frac{t}{20} \\ t & = 80 \min \end{aligned}$

2. Given below are two statements

Statement I : The law of radioactive decay states that the number of nuclei undergoing the decay per unit time is inversely proportional to the total number of nuclei in the sample.

Statement II : The half of a radionuclide is the sum of the life time of all nuclei, divided by the initial concentration of the nuclei at time t = 0.

In the light of the above statements, choose the most appropriate answer from the options given below : ⇒ ( NEET 2022 Phase 2)

A. Statement I is incorrect but statement II is correct

B. Both statement I and statement II are correct

C. Both statement I and statement II are incorrect

D. Statement I is correct but statement II is incorrect

The correct answer is option (C)

According to law of radioactive decay

Rate of decay $(- \frac{d N}{d t}) \propto N$ (No. of undecayed nuclei)

Half life is the time in which half of the radioactive sample decays

$∴$ Both statements are incorrect.

3. At any instant, two elements X₁ and X₂ have same number of radioactive atoms. If the decay constant of X₁ and X₂ are 10 $λ$ and $λ$ respectively, then the time when the ratio of their atoms becomes $\frac{1}{e}$ respectively will be : ⇒ (NEET 2022 Phase 2)

A. $\frac{1}{5 λ}$

B. $\frac{1}{11 λ}$

C. $\frac{1}{9 λ}$

D. $\frac{1}{6 λ}$

The correct answer is option (C)

Number of radioactive nuclei at any time is

$N = N_{0} e^{- λ t}$

Initial number of nuclei is same for sample X₁ & X₂.

Let after time 't' the ratio of their atoms become $\frac{1}{e}$

$\Rightarrow \frac{N_{0} e^{- 10 λ \times t}}{N_{0} e^{- λ \times t}} = \frac{1}{e} \Rightarrow e^{- 9 λ t} = e^{- 1} \Rightarrow t = \frac{1}{9 λ}$

Hence, $t = \frac{1}{9 λ}$

4. The half life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be ⇒ (NEET 2021)

A. $\frac{2}{3 \sqrt{2}}$

B. $\frac{1}{2}$

C. $\frac{1}{2 \sqrt{2}}$

D. $\frac{2}{3}$

The correct answer is option (C)

$\frac{A}{A_{0}} = {(\frac{1}{2})}^{t / T_{H}} = {(\frac{1}{2})}^{150 / 100} = \frac{1}{2 \sqrt{2}}$

5. For a radioactive material, half-life is 10 minutes. If initially there are 600 number of nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is ⇒ (NEET 2018)

A. 20

B. 10

C. 30

D. 15

The Correct Answer is Option (A)

Given N₀ = 600, N₁ = 450, T = 10 min

Number of nuclei remaining, N = 600 – 450 = 150 after time ‘t’

$\frac{N}{N_{0}} = \frac{150}{600} = \frac{1}{4}$

According to the law of radioactive decay,

N = N₀e^{- $λ$ t}

$\frac{N}{N_{0}} = {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$

$∴$ $\frac{150}{600} = {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$

$\Rightarrow$ ${(\frac{1}{2})}^{2} = {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$

$\Rightarrow$ t = 2T_1/2 = 2 $\times$ 10 minutes = 20 minutes

6. Radioactive material 'A' has decay constant '8 $λ$ ' and material 'B' has decay constant ' $λ$ '. Initially they have same number of nuclei. After what time, the ratio of number of nuclei of material 'B' to that 'A' will be e ? ⇒ ()

A. $\frac{1}{7 λ}$

B. $\frac{1}{8 λ}$

C. $\frac{1}{9 λ}$

D. $\frac{1}{λ}$

The Correct Answer is Option ( A)

Ratio of number of nuclei

$\frac{N_{1}}{N_{2}} = \frac{e^{- 8 λ t}}{e^{- λ t}}$ = $e^{- 7 λ t}$

According to question,

$e^{- 7 λ t}$ = $\frac{1}{e}$

$\Rightarrow$ t = $\frac{1}{7 λ}$

7. The half-life of a radioactive substance is 30 minutes. The time (in minutes) taken between 40% decay and 85% decay of the same radioactive substance is⇒ (NEET 2016 Phase 2)

A. 15

B. 30

C. 45

D. 60

The Correct Answer is Option (D)

N₀ = Nuclei at time t = 0

N₁ = Remaining nuclei after 40% decay

= (1 – 0.4) N₀ = 0.6 N₀

N₂ = Remaining nuclei after 85% decay

= (1 – 0.85) N₀ = 0.15 N₀

$∴$ $\frac{N_{2}}{N_{1}}$ = $\frac{0.15 N_{0}}{0.6 N_{0}}$ = ${(\frac{1}{2})}^{2}$

Hence number of half-lives is 2, so the time needed is :

t = 2 × 30 = 60 mints

8. A radioactive X with a half life 1.4 $\times$ 10⁹ years decays to Y which is stable. A sample of the rock from a cave was found to contain X and Y in the ratio 1 : 7. The age of the rock is ⇒ (AIPMT 2014)

A. 1.96 $\times$ 10⁹ years

B. 3.92 $\times$ 10⁹ years

C. 4.20 $\times$ 10⁹ years

D. 8.40 $\times$ 10⁹ years

The Correct Answer is Option (C)

Using N = N₀e^–λt

Now 1 = 8e ^–λt

or, $\frac{1}{8}$ = e^{– (ln 2/T)t}

or, ${(\frac{1}{2})}^{3}$ = (2)^–t/T

or, t/T = 3

Now time, t = 3 × 1.4 × 10⁹ = 4.2 × 10⁹ years

9. The half life of a radioactive isotope 'X' is 20 years. It decays to another element 'Y' which is stable. The two elements 'X' and 'Y' were found to be in the ratio 1 : 7 in a sample of a given rock. The age of the rock is estimated to be ⇒ (NEET 2013)

A. 80 years

B. 100 years

C. 40 years

D. 60 years

The Correct Answer is Option (D)

Initial number of atoms of X is N₀

and Initial number of atoms of Y is 0

Number of atoms after time t, for X is N and for Y is N₀ - N

According to the question,

$\frac{N}{N_{0} - N}$ = $\frac{1}{7}$

$\Rightarrow$ $\frac{N}{N_{0}} = \frac{1}{8}$

As $\frac{N}{N_{0}} = {(\frac{1}{2})}^{n}$ where n is the no. of half lives

$∴$ ${(\frac{1}{2})}^{n} = \frac{1}{8}$

$\Rightarrow$ n = 3

t = nT_1/2 = 3 × 20 = 60 years

Hence, the age of rock is 60 years.

10. The half life of a radioactive nucleus is 50 days. The time invertal (t₂ $-$ t₁) between the time t₂ when $\frac{2}{3}$ of it has decayed and the time t₁ when $\frac{1}{3}$ of it had decayed is ⇒ (AIPMT 2012 Mains)

A. 30 days

B. 50 days

C. 60 days

D. 15 days

The Correct Answer is Option (B)

At time t₂, $\frac{2}{3}$ of the sample had decayed

$∴$ N = $\frac{1}{3}$ N₀

$∴$ $\frac{1}{3}$ N₀ = N₀ $e^{- λ t_{2}}$ ......(1)

At time t₁, $\frac{1}{3}$ of the sample had decayed

$∴$ N = $\frac{2}{3}$ N₀

$∴$ $\frac{2}{3}$ N₀ = N₀ $e^{- λ t_{1}}$ ......(2)

Divide (i) by (ii), we get

$\frac{1}{2} = \frac{e^{- λ t_{2}}}{e^{- λ t_{1}}}$

$\Rightarrow$ $\frac{1}{2} = e^{- λ (t_{2} - t_{1})}$

$\Rightarrow$ $λ (t_{2} - t_{1})$ = ln 2

$\Rightarrow$ $(t_{2} - t_{1})$ = $\frac{\ln 2}{λ}$ = $\frac{\ln 2}{\frac{\ln 2}{T_{1 / 2}}}$ = T_1/2 = 50 days

11. A mixture consists of two radioactive materials A₁ and A₂ with half lives of 20 s and 10 s respectively. Initially the mixture has 40 g of A₁ and 160 g of A₂. The amount of the two in the mixture will become equal after ⇒ (AIPMT 2012 Prelims)

A. 60 s

B. 80 s

C. 20 s

D. 40 s

The Correct Answer is Option (D)

Let, the amount of the two in the mixture will become equal after t years.

The amount of A₁, which remains after t years

N₁ = $\frac{N_{01}}{{(2)}^{t / 20}}$

The amount of A₂, which remains, after t years

N₂ = $\frac{N_{02}}{{(2)}^{t / 10}}$

According to the question

N₁ = N₂

$\Rightarrow$ $\frac{40}{{(2)}^{t / 20}} = \frac{160}{{(2)}^{t / 10}}$

$\Rightarrow$ ${(2)}^{t / 20}$ = ${(2)}^{(\frac{t}{10} - 2)}$

$\Rightarrow$ $\frac{t}{20} = (\frac{t}{10} - 2)$

$\Rightarrow$ t = 40 s

12. Two radioactive nuclei P and Q, in a given sample decay into a stable nucleus R. At time t = 0. number of P species are 4 N₀ and that of Q are N₀. Half -life of P (for conversion to R) is 1 minute where as that of Q is 2 minutes. Initially there are no nuclei of R present in the sample. When number of nuclei of P and Q are equal, the number of nuclei of R present in the sample would be ⇒ (AIPMT 2011 Mains)

A. 2 N₀

B. 3 N₀

C. $\frac{9 N_{0}}{2}$

D. $\frac{5 N_{0}}{2}$

The Correct Answer is Option (C)

Initially P = 4N₀

Q = N₀

Half life T_P = 1 min.

T_Q = 2 min.

Let after time t number of nuclei of P and Q are equal, that is

N_P = N_Q

$\Rightarrow$ $4 N_{0} {(\frac{1}{2})}^{t / 1}$ = $N_{0} {(\frac{1}{2})}^{t / 2}$

$\Rightarrow$ $2^{t / 1} = {4.2}^{t / 2}$

$\Rightarrow$ t = 4 min

After 4 minutes, both P and Q have equal number of nuclei.

$∴$ Number of nuclei of R

= $(4 N_{0} - \frac{N_{0}}{4})$ + $(N_{0} - \frac{N_{0}}{4})$

= $\frac{9 N_{0}}{2}$

⇒ Nuclei

Topic 4 : Law of Radioactive Decay