Radioactivity Multiple Choice Questions And Answers

Radioactivity

Question 1. The half-life of a radioactive material is 3 h. If the initial amount is 300 g then after 18 h it will remain

  1. 4.68 g
  2. 9.375 g
  3. 48.8 g
  4. 93.75 g

Answer: 1. 4.68 g

Given that total time = t =18 h and half-life = T1/2 = 3 h.

∴ the number of half-lives is

⇒ \(n=\frac{t}{T_{1 / 2}}=\frac{18 \mathrm{~h}}{3 \mathrm{~h}}=6\)

Now, the amount of active material left after 6 half-lives is

∴ \(N=\left(\frac{1}{2}\right)^6 N_0=\frac{300 \mathrm{~g}}{64}=4.68 \mathrm{~g}\)

Alternative method

⇒ \(\begin{aligned}
300 \mathrm{~g} \stackrel{T_{1 / 2}}{\longrightarrow} 150 \mathrm{~g} \stackrel{T_{1 / 2}}{\longrightarrow} 75 \mathrm{~g} \stackrel{T_{1 / 2}}{\longrightarrow} 37.5 \mathrm{~g} \\
\stackrel{T_{1 / 2}}{\longrightarrow} 18.75 \mathrm{~g} \stackrel{T_{1 / 2}}{\longrightarrow} 9.37 \mathrm{~g} \stackrel{T_{1 / 2}}{\longrightarrow} 4.68 \mathrm{~g}
\end{aligned}\)

After 6 half-lives, 4.68 g of active nuclei remains

Question 2. The half-lives of two radioactive elements A and B are respectively 20 min and 40 min. Initially, the samples of A and B have equal numbers of nuclei. After 80 minutes, the ratio of the remaining numbers of the A and B nuclei will be

  1. 1:16
  2. 1:4
  3. 1 :1
  4. 4: 1

Answer: 2. 1:4

For A, half-life = TA = 20 min, and for B, half-life = TB = 40 min.

Total time = 80 min = 4TA = 2TB

∴ \(N_{\mathrm{A}}=\left(\frac{1}{2}\right)^4 N_0 \text { and } N_{\mathrm{B}}=\left(\frac{1}{2}\right)^2 N_0\)

Hence, the ratio between the numbers of active nuclei of A and B is

∴ \(\frac{N_{\mathrm{A}}}{N_{\mathrm{B}}}=\frac{\left(\frac{1}{2}\right)^4}{\left(\frac{1}{2}\right)^2}=\frac{4}{16}=\frac{1}{4} \Rightarrow N_{\mathrm{A}}: N_{\mathrm{B}}=1: 4\)

Question 3. The half-life of radium is 1600 y. The fraction of a sample of radium that would remain active after 6400 y is

  1. \(\frac{1}{4}\)
  2. \(\frac{1}{2}\)
  3. \(\frac{1}{8}\)
  4. \(\frac{1}{16}\)

Answer: 4. \(\frac{1}{16}\)

For radium, half-life = T1/2 = 1600 y.

Total time = t = 6400 y = 4T1/2.

⇒ \(N=\left(\frac{1}{2}\right)^4 N_0\)

Hence, the fraction of nuclei which remain active will be

∴ \(\frac{N}{N_0}=\left(\frac{1}{2}\right)^4=\frac{1}{16}\)

Question 4. The activity of a radioactive sample is measured as 9750 counts min-1 at t = 0 and 975 counts min-1 at t = 5 min. The decay constant is approximately

  1. 0.92 mm-1
  2. 0.69 min-1
  3. 0.46 min-1
  4. 0.23 min-1

Answer: 3. 0.46 min-1

Activity (A) is measured as the rate of disintegration.

∵ \(N=N_0 \mathrm{e}^{-\lambda .}\)

∴ \(A=\left|\frac{d N}{d t}\right|=N_0 \lambda \mathrm{e}^{-\lambda t}=A_0 \mathrm{e}^{-\lambda t}\)

where AQ is the activity at t = 0.

Given that A0 = 9750 counts min-1 at t = 0, and A = 975 counts min-1 at t = 5 min.

∴ \(A=A_0 \mathrm{e}^{-\lambda t} \Rightarrow \frac{A_0}{A}=\frac{9750}{975}=\mathrm{e}^{\lambda t}\)

∴ In 10 = λ(5 min)

∴ decay constant = \(\lambda=\frac{\ln 10}{5 \min }=\frac{2.3}{5 \mathrm{~min}}=0.46 \mathrm{~min}^{-1}\)

Question 5. The relation between disintegration constant (λ) and half-life (T1/2) is

  1. \(\lambda=\frac{\log _{10} 2}{T}\)
  2. \(\lambda=\frac{\log _2 \mathrm{e}}{T}\)
  3. \(\lambda=\frac{\log _{\mathrm{e}} 2}{T}\)
  4. \(\lambda=\frac{\log _{\mathrm{e}} T_{1 / 2}}{2}\)

Answer: 3. \(\lambda=\frac{\log _{\mathrm{e}} 2}{T}\)

Since the radioactive decay follows the exponential law, N = N0e-λt.

After one half-life, N0 changes to N0/2

⇒ \(\frac{N_0}{2}=N_0 \mathrm{e}^{-\lambda T} \Rightarrow 2=\mathrm{e}^{\lambda T}\)

∴ \(\lambda T=\ln 2 \Rightarrow \lambda=\frac{\ln 2}{T}=\frac{\log _e^2}{T}\)

Question 6. A sample of a radioactive element contains 4 x 1010 active nuclei. If the half-life of the element is 10 days, the number of decayed nuclei after 30 days is

  1. 0.5 x l010
  2. 2 x l010
  3. 3.5 x l010
  4. 1 x l010

Answer: 3. 3.5 x l010

Given that N0 = 4 x l010, half-life = T1/2=10 days, and total time = t = 30 days.

number of half-lives = n = \(\frac{t}{T_{1 / 2}}=\frac{30 \text { days }}{10 \text { days }}=3\)

Hence, the number of active nuclei after 3 half-lives is

⇒ \(N=\left(\frac{1}{2}\right)^3 N_0=\frac{N_0}{8}\)

∴ the number of decayed nuclei is

∴ \(N_0-N=N_0\left(1-\frac{1}{8}\right)=\frac{7}{8} N_0=\frac{7}{8}\left(4 \times 10^{10}\right)=3.5 \times 10^{10}\).

Question 7. The half-life of a radioactive sample is 6 h. After 24 h, its activity is 0.01 μCi. What was the initial activity?

  1. 0.04 Ci
  2. 0.08 μCi
  3. 0.16 μCi
  4. 0.24 Ci

Answer: 3. 0.16 μCi

The activity at the time t is

⇒ \(A(t)=A_0 \mathrm{e}^{-\lambda t}\),

where A0 = initial activity.

Given that half-life = T1/2 = 6 h and total time = t = 24 h.

The activity at the time t is 0.01 μCi.

∴ \(0.01 \mu \mathrm{Ci}=A_0 \mathrm{e}^{-(\ln 2)(t / T)}\)

∴ initial activity = \(A_0=(0.01 \mu \mathrm{Ci}) \mathrm{e}^{(\ln 2)(24 / 6)}\)

= (0.01 (μCi)e41n 2 = (0.01 μCi)(16) = 0.16 μCi.

Question 8. The half-life of radium is 1600 y. What fraction of the same remains undecayed after four half-lives?

  1. \(\frac{1}{8}\)
  2. \(\frac{1}{4}\)
  3. \(\frac{1}{16}\)
  4. \(\frac{1}{24}\)

Answer: 3. \(\frac{1}{16}\)

Given that half-life = T1/2 = 1600 y and total time = t = 4T1/2 = 4 half-lives.

∴ the number of undecayed (active) nuclei after 4 half-lives will be

⇒ \(N=\left(\frac{1}{2}\right)^4 N_0=\frac{N_0}{16}\)

∴ required fraction = \(\frac{N}{N_0}=\frac{1}{16}\)

Question 9. M alpha particles per second are emitted from N atoms of a radioactive element. The half-life of the radioactive element is

  1. \(\frac{M}{N} \mathrm{~s}\)
  2. \(\frac{N}{M} \mathrm{~s}\)
  3. \(\frac{N(0.693)}{M} \mathrm{~s}\)
  4. \(\frac{M(0.693)}{N} \mathrm{~s}\)

Answer: 3. \(\frac{N(0.693)}{M} \mathrm{~s}\)

For a nuclear disintegration,

⇒ \(N=N_0 \mathrm{e}^{-\lambda t} \text { and } A=\frac{d N}{d t}=\lambda N_0 \mathrm{e}^{-\lambda t}=\lambda N\)

Given that disintegration rate = A = M s-1

∴\(M=\lambda N \Rightarrow \lambda=\frac{M}{N}\)

∴ half life \(T_{1 / 2}=\frac{\ln 2}{\lambda}=\frac{\ln 2}{M / N} \mathrm{~s}=\frac{N}{M}(0.693) \mathrm{s}\)

Question 10. The half-life of a radioactive element is 5 y. What percentage of it will remain active after 25 y?

  1. 25%
  2. 6.25%
  3. 1.25%
  4. 3.125%

Answer: 4. 3.125%

Given that half-life = T1/2 = 5 y and total time = t = 25 y = 5(5 y) = 5T1/2.

After 5 half-lives, the number of active (undecayed) nuclei will be

⇒ \(N=\left(\frac{1}{2}\right)^5 N_0=\frac{N_0}{32}\)

∴ \(\frac{N}{N_0}=\frac{1}{32}=\frac{100 \%}{32}=3.125 \%\)

Question 11. Two radioactive substances A and B have their decay constants 5λ. and A respectively. At t = 0, they have the same number of nuclei. The ratio of the number of active nuclei of A to those of B will be 1/e2 after a time period of

  1. \(\frac{1}{2 \lambda}\)
  2. \(\frac{1}{4 \lambda}\)

Answer: 3. \(\frac{1}{2 \lambda}\)

The number of active nuclei of element A is NA = N0e-5λt and that of element B is NB = N0e-λt.

Given that \(\frac{N_{\mathrm{A}}}{N_{\mathrm{B}}}=\frac{1}{\mathrm{e}^2}=\frac{\mathrm{e}^{-5 \lambda t}}{\mathrm{e}^{-\lambda t}}=\frac{1}{\mathrm{e}^{4 \lambda t}}\)

4λt = 2.

Hence, the required time is t = \(t=\frac{1}{2 \lambda}\)

Question 12. Nl atoms of a radioactive element emit N2 beta particles per second. The decay constant of the element is

  1. \(\frac{N_1}{N_2}\)
  2. \(\frac{N_2}{N_1}\)
  3. N1 In 2
  4. N2 In 2

Answer: 2. \(\frac{N_2}{N_1}\)

Activity = \(A=\left|\frac{d N}{d t}\right|=\lambda N_0 \mathrm{e}^{-\lambda t}=\lambda N\)  → (1)

Since N2 beta particles are emitted from N1 nuclei, we have

A = N2 S-1 and N = N1

From (1), N2= N1.

Hence, decay constant = \(=\lambda=\frac{N_2}{N_1}\)

Question 13. The half-life of a radioactive substance is 30 min. The time taken between the 40% decay and the 85% decay of this radioactive substance is

  1. 45 min
  2. 15 min
  3. 60 min
  4. 30 min

Answer: 3. 60 min

∵ 40% decay = 60% active nuclei,

∴ \(N_1=\frac{60}{100} N_0\)

∵ 85% decay =15% active nuclei,

∴ \(N_2=\frac{15}{100} N_0\)

∴ \(\frac{N_2}{N_1}=\frac{1}{4}=\left(\frac{1}{2}\right)^2\)

Thus, the time duration is 2 half-lives = 2(30 min) = 60 min

Question 14. A mixture consists of two radioactive substances A1 and A2 having the half-lives 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 a period of

  1. 60 s
  2. 80 s
  3. 40 s
  4. 20 s

Answer: 3.

For A1, half-life = 20 s.

For A2, half-life = 10 s.

Let A1 be reduced to N from N0 in the time t = (t/T1/2) half-lives

∴ \(N=N_0\left(\frac{1}{2}\right)^{\frac{t}{T_{1 / 2}}}=(40 \mathrm{~g})\left(\frac{1}{2}\right)^{t / 20 \mathrm{~s}}\)

Similarly, for A2,

⇒ \(N=(160 \mathrm{~g})\left(\frac{1}{2}\right)^{t / 10 \mathrm{~s}}\)

∴ \(\left(\frac{1}{2}\right)^{t / 20 s}=4\left(\frac{1}{2}\right)^{t / 10 s}\)

⇒ \(\frac{t}{20 \mathrm{~s}}+2=\frac{t}{10 \mathrm{~s}}\)

⇒ \(\frac{t}{20 \mathrm{~s}}=2 \Rightarrow t=40 \mathrm{~s}\)

Question 15. A sample of a radioactive element has a mass of 10 g at the instant t = 0. The approximate mass of the element in the sample after two mean lives is

  1. 3.70 g
  2. 1.35 g
  3. 6.30 g
  4. 2.50 g

Answer: 2. 1.35 g

At f = 0, M0 = 10 g.

Mean life = \(\tau=\frac{1}{\lambda}\)

∴ the active mass after 2 mean lives (= 2/λ) will be

∴ \(M=M_0 \mathrm{e}^{-\lambda(2 / \lambda)}=M_0 \mathrm{e}^{-2}=\frac{10 \mathrm{~g}}{\mathrm{e}^2}=\frac{10 \mathrm{~g}}{7.39}=1.35 \mathrm{~g}\)

Question 16. If the half-lives of a radioactive substance for its α- and β -decays are 4 y and 12 y respectively, the ratio of the initial activity to that after 12 y will be

  1. 6.25%
  2. 12.5%
  3. 25%
  4. 50%

Answer: 1. 6.25%

For the α-decay, Tα= 4 y.

For the β-decay, Tβ =12 y

∴ net activity = \(\frac{d N_{\mathrm{A}}}{d t}=\frac{d N_{\mathrm{B}}}{d t}+\frac{d N_{\mathrm{C}}}{d t}\)

∴ \(\lambda_{\text {eff }} N_{\mathrm{A}}=\lambda_1 N_{\mathrm{A}}+\lambda_2 N_{\mathrm{A}}\)

⇒ \(\lambda_{\text {eff }}=\lambda_1+\lambda_2\)

⇒ \(\frac{1}{\lambda_{\text {eff }}}=\frac{1}{T_1}+\frac{1}{T_2}=\frac{T_1+T_2}{T_1 T_2}\)

∴ effective half-life = \(T_{\text {eff }}=\frac{T_1 T_2}{T_1+T_2}=\frac{4 \times 12}{16} \mathrm{y}=3 \mathrm{y}\)

∴ number of half-lives = n = \(n=\frac{12 y}{3 y}=4\)

∴ \(\frac{N}{N_0}=\left(\frac{1}{2}\right)^n=\left(\frac{1}{2}\right)^4=\frac{1}{16}=6.25 \%\)

Radioactivity Multiple Choice Questions And Answers Q16

Question 17. The half-life of a substance is 20 min. What is the time interval between its 33% decay and 67% decay?

  1. 40 min
  2. 25 min
  3. 20 min
  4. 30 min

Answer: 3. 20 min

Given that half-life = T1/2 = 20 min.

Assume that at a time t1, undecayed nuclei = (100- 33)%.

∴ N1 = (67%)N0.

Similarly, assume that at a time t2,

N2 = (100- 67)%. N0 = (33%)N0.

∴ \(\frac{N_1}{N_2}=\frac{N_0 \mathrm{e}^{-\lambda t_1}}{N_0 \mathrm{e}^{-\lambda t_2}} \Rightarrow \frac{67}{33}=\mathrm{e}^{\lambda\left(t_2-t_1\right)}\)

Taking \(\frac{67}{33}=2\), we have

⇒ \(2=\mathrm{e}^{\left(t_2-t_1\right) \ln 2 / T}=\mathrm{e}^{\ln 2^{\Delta t / T}}=2^{\Delta t / T}\)

∴ \(\frac{\Delta t}{T}=1\).

Hence, the required time interval is Δt = T1/2 = one half-life = 20 min

Question 18. A radioisotope X with a half-life of 1.4 x l09 y decays to Y, which is a stable isotope. A sample of a rock from a cave was found to contain X and Y in a ratio 1:7. The age of the rock is

  1. 1.96 x l09 y
  2. 3.92 x l09 y
  3. 8.40 x l09 y
  4. 4.20 x l09 y

Answer: 4. 4.20 x l09 y

Let us assume that at t = 0, the total number of active X nuclei = N0 which decay to Y nuclei such that Nx : NY =1: 7.

∴ \(N_{\mathrm{X}}=\frac{N_0}{8} \text { and } N_{\mathrm{Y}}=\frac{7 N_0}{8}\)

Now, \(N_{\mathrm{X}}=\frac{N_0}{8}=\left(\frac{1}{2}\right)^3 N_0\)

This occurs after 3 half-lives. Thus,

required time = t = \(3 T_{1 / 2}=3\left(1.4 \times 10^9 \mathrm{y}\right)=4.2 \times 10^9 \mathrm{y}\).

Alternative method

⇒ \(N_{\mathrm{X}}=\frac{N_0}{8}=N_0 \mathrm{e}^{-\lambda t}\)

⇒ \(8=2^3=\mathrm{e}^{\lambda t}\)

∴ \(\ln 2^3=\ln \mathrm{e}^{\lambda t} \Rightarrow 3 \ln 2=\lambda t\)

⇒ \(t=3\left(\frac{\ln 2}{\lambda}\right)=3 T=3\left(1.4 \times 10^9 \mathrm{y}\right)=4.2 \times 10^9 \mathrm{y}\)

Question 19. A sample of a radioactive material decays at the rate of 500 per second at a certain time. The activity falls to 200 per second after 50 min. What is the decay constant (λ) of the sample? (Given that In 2.5 = 0.916.)

  1. 6.10 x l0-4 s-1
  2. 3.05 x l0-4 s-1
  3. 5.0 x 10-4 s-1
  4. 1.0 x l0-4 s-1

Answer: 2. 3.05 x l0-4 s-1

Initial decay rate = A0 = 500 s-1.

⇒ A(t) = A0e-λt

⇒ 200 s-1 = (500 s-1 )e(5° min)

2.5 = eλ(50 min)

∴ In 2.5 = In eλ(50 min)

⇒ In 2.5 = λ(50 min).

Hence, decay constant = \(\frac{\ln 2.5}{50 \times 60 \mathrm{~s}}=\frac{0.916}{30} \times 10^{-2} \mathrm{~s}^{-1}\)

= \(3.05 \times 10^{-1} \mathrm{~s}^{-1}\).

20. In a radioactive material, the activity at a time t1 is R1, and at a later time t2, it is R2. If the decay constant of the material is λ then

  1. \(R_1=R_2 \mathrm{e}^{-\lambda\left(t_1-t_2\right)}\)
  2. \(R_1=R_2 \mathrm{e}^{\lambda\left(t_1-t_2\right)}\)
  3. \(R_1=R_2\left(\frac{t_2}{t_1}\right)\)
  4. R1 = R2

Answer: 1. \(R_1=R_2 \mathrm{e}^{-\lambda\left(t_1-t_2\right)}\)

We know that A(t) = A0e-λt.

∴ at \(t=t_1, R_1=R_0 \mathrm{e}^{-\lambda t_1} \text { and at } t=t_2, R_2=R_0 \mathrm{e}^{-\lambda t_2}\)

Hence, \(\frac{R_1}{R_2}=\mathrm{e}^{-\lambda\left(t_1-t_2\right)}\)

⇒ \(R_1=R_2 \mathrm{e}^{-\lambda\left(t_1-t_2\right)}\).

21. The half-life of a radioactive isotope X is 20 y. 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

  1. 40 y
  2. 100 y
  3. 60 y
  4. 80 y

Answer: 3. 60 y

The solution is similar to that of Q.18.

Only the half-life is changed to T1/2 = 20 y.

Hence, the age of the rock will be 3 half-lives, i.e., T1/2 = 3(20 y) = 60 y

22. The activity of a radioactive sample is measured as N0 counts per minute at t = 0 and N0/e counts per minute at t = 5 min. The time period at which the activity reduces to half its value is

  1. \(\ln \frac{2}{5} \min\)
  2. \(\frac{5}{\ln 2} \min\)
  3. \(5 \log 2 \mathrm{~min}\)
  4. \(5 \ln 2 \mathrm{~min}\)

Answer: 4. \(5 \ln 2 \mathrm{~min}\)

At t = 0, activity = N0.

At t = 5 min,

activity = \(\frac{N_0}{\mathrm{e}}=N_0 \mathrm{e}^{-\lambda(5 \mathrm{~min})}\)

⇒ \(\lambda(5 \mathrm{~min})=1 \Rightarrow \lambda=\frac{1}{5 \mathrm{~min}}\)

When the activity reduces to half, the time taken is one half-life. Thus,

∴ \(t=T_{1 / 2}=\frac{\ln 2}{\lambda}=\frac{\ln 2}{1 / 5 \mathrm{~min}}=(5 \ln 2) \mathrm{min}\).

Question 23. A radioactive isotope is being produced at a constant rate α per second. Its decay constant is λ. If N0 is the number of nuclei at a time t = 0, the maximum possible number of nuclei is

  1. \(N_0+\frac{\alpha}{\lambda}\)
  2. \(N_0\)
  3. \(\frac{\lambda}{\alpha}+N_0\)
  4. \(\frac{\alpha}{\lambda}\)

Answer: 4. \(\frac{\alpha}{\lambda}\)

The number of nuclei present will be maximum when the decay rate is equal to the rate at which isotopes are being produced.

Now, rate of decay = \(=\left|\frac{d N}{d t}\right|=N_0 \lambda e^{-\lambda t}=\lambda N\) and rate of formation = a.

∴ λN = α

Hence, the maximum number of nuclei is N = α/λ.

Question 24. If the half-life of a radioactive sample is 69.3 h, how much of the initial number will decay between the tenth and eleventh hours?

  1. 1%
  2. 2%
  3. 3%
  4. 4%

Answer: 1. 1%

The number of active nuclei at the time t1 = 10 h is N1 = N0e-10λ, and that at the time t2 =11 h is N2 = N0e-11λ.

∴ percentage decay = \(\frac{N_1-N_2}{N_1} \times 100 \%\)

⇒ \(\frac{N_0 \mathrm{e}^{-10 \lambda}-N_0 \mathrm{e}^{-11 \lambda}}{N_0 \mathrm{e}^{-10 \lambda}} \times 100 \%=\left(1-\frac{1}{e^\lambda}\right) \times 100 \%\)

Given that half-life = \(T_{1 / 2}=69.3 \mathrm{~h}=\frac{\ln 2}{\lambda}=\frac{0.693}{\lambda}\)

∴ decay constant = λ = 0.01 h-1 .

∴ \(\% \text { decay }=\left(1-\frac{1}{\mathrm{e}^{0.01}}\right) \times 100 \%=(1-0.99) \times 100 \%=1 \%\)

Question 25. Two radioactive materials A and B have their decay constants 10λ and λ respectively. If initially they have the same number of nuclei, the ratio of the number of nuclei of A and B will be 1/e after a time period of

  1. \(\frac{1}{9 \lambda}\)
  2. \(\frac{1}{11 \lambda}\)
  3. \(\frac{1}{10 \lambda}\)
  4. \(\frac{11}{10 \lambda}\)

Answer: 1. \(\frac{1}{9 \lambda}\)

For the element A, \(N_{\mathrm{A}}=N_0 \mathrm{e}^{-10 \lambda t}\)

For the element B, \(N_{\mathrm{B}}=N_0 \mathrm{e}^{-\lambda t}\)

∴ \(\frac{N_{\mathrm{A}}}{N_{\mathrm{B}}}=\frac{1}{\mathrm{e}}=\frac{\mathrm{e}^{-10 \lambda t}}{\mathrm{e}^{-\lambda t}}=\frac{1}{\mathrm{e}^{9 \lambda t}}\)

∴ \(9 \lambda t=1 \Rightarrow t=\frac{1}{9 \lambda}\)

Question 26. A sample of a radioactive material A that has an activity of 10 mCi (given that 1 Ci = 3.7 x 1010 decays per second) has twice the number of nuclei as another sample of a different radioactive material B, which has an activity of 20 mCi. The correct choices for the half-lives of A and B would then be respectively

  1. 20 d and 10 d
  2. 5 d and 10 d
  3. 10 d and 40 d
  4. 20 d and 5 d

Answer: 4. 20 d and 5 d

The activity (A) of a radioactive sample is the rate of disintegration. So,

⇒ \(A=\left|\frac{d N}{d t}\right|=\lambda N=\left(\frac{\ln 2}{T}\right) N\),

where T is the half-life of the material.

∴ \(\frac{A_{\mathrm{A}}}{A_{\mathrm{B}}}=\frac{N_{\mathrm{A}}}{N_{\mathrm{B}}} \cdot \frac{T_{\mathrm{B}}}{T_{\mathrm{A}}} \Rightarrow \frac{10 \mathrm{mCi}}{20 \mathrm{mCi}}=\frac{2 N_{\mathrm{B}}}{N_{\mathrm{B}}} \cdot \frac{T_{\mathrm{B}}}{T_{\mathrm{A}}}\)

∴ \(\frac{T_{\mathrm{A}}}{T_{\mathrm{B}}}=\frac{4}{1}\)

This is true for the option (4).

Question 27. Using a nuclear counter, the count rate of particles emitted from a radioactive source is measured. At t = 0, it was 1600 per second, and at t = 8 s, it was 100 per second. The count rate observed at t = 6 s is close to

  1. 400 s-1
  2. 200 s-1
  3. 150 s-1
  4. 300 s-1

Answer: 2. 200 s-1

The count rate is the same as the activity, i.e., A = \(A=\left|\frac{d N}{d t}\right|\)

But \(N=N_0 \mathrm{e}^{-\lambda t}\)

So, \(A=\left|\frac{d N}{d t}\right|=\lambda N_0 \mathrm{e}^{-\lambda t}\)

A(t) = A(0)e-λt.

Given that A(0) = 1600 s-1 and A(8) = 100s-1.

⇒ \(\frac{A(0)}{A(8)}=\frac{1600}{100}=\mathrm{e}^{\lambda .8} \Rightarrow 2^4=\left(\mathrm{e}^{2 \lambda}\right)^4\)

⇒ \(\mathrm{e}^{2 \lambda}=2 \Rightarrow \lambda=\frac{1}{2} \ln 2=\ln \sqrt{2}\)

At t = 6 s,

∴ \(A(t)=A(6 \mathrm{~s})=N_0 \mathrm{e}^{-6 \lambda}=\left(1600 \mathrm{~s}^{-1}\right) \mathrm{e}^{-6 \ln \sqrt{2}}=\frac{1600 \mathrm{~s}^{-1}}{8}=200 \mathrm{~s}^{-1}\).

Question 28. Two radioactive samples A and B have their decay constants 5λ and λ respectively. At a time t = 0, the samples have equal numbers of active nuclei. The time taken for the ratio of the numbers of nuclei to become 1/e2 will be

  1. \(\frac{2}{\lambda}\)
  2. \(\frac{1}{\lambda}\)
  3. \(\frac{1}{4 \lambda}\)
  4. \(\frac{1}{2 \lambda}\)

Answer: 4. \(\frac{1}{2 \lambda}\)

⇒ \(\frac{N_{\mathrm{A}}}{N_{\mathrm{B}}}=\frac{N_0 \mathrm{e}^{-5 i t}}{N_0 \mathrm{e}^{-\lambda, t}}=\mathrm{e}^{-4 i t}\) → (1)

Bur \(\frac{N_{\mathrm{A}}}{N_{\mathrm{B}}}=\frac{1}{\mathrm{e}^2}=\mathrm{e}^{-2}\) → (2)

From (1) and (2)

∴ \(4 \lambda t=2 \Rightarrow t=\frac{1}{2 \lambda}\)

Question 29. The half-lives of two radioactive samples A and B are 10 min and 20 min respectively. If initially a sample has equal numbers of the nuclei of A and B then after 60 min the ratio of the decayed numbers of the nuclei A and B will be

  1. 1: 8
  2. 9: 8
  3. 8: 1
  4. 3: 8

Answer: 2. 9: 8

The number of active nuclei at a time t is N = \(N_0 \mathrm{e}^{-\lambda t}\)

⇒ \(\Delta N=N_0-N=N_0\left(1-\mathrm{e}^{-\lambda t}\right)\)

∴ \(\frac{\Delta N_{\mathrm{A}}}{\Delta N_{\mathrm{B}}}=\frac{1-\mathrm{e}^{-\lambda_{\mathrm{A}} t}}{1-\mathrm{e}^{-\lambda_{\mathrm{B}} t}}=\frac{1-\mathrm{e}^{-\left[(\ln 2) / T_{\mathrm{A}}\right] t}}{1-\mathrm{e}^{-\left[(\ln 2) / T_{\mathrm{B}}\right] t}}\)

⇒ \(\frac{1-\mathrm{e}^{-[(\ln 2) / 10] \times 60}}{1-\mathrm{e}^{-[(\ln 2) / 20] \times 60}}=\frac{1-\frac{1}{2^6}}{1-\frac{1}{2^3}}=\frac{63}{64} \times \frac{8}{7}=\frac{9}{8}\)

∴ \(\Delta N_{\mathrm{A}}: \Delta N_{\mathrm{B}}=9: 8\).

Question 30. At a time t = 0, the activity of two radioactive elements A and B are equal. After a time t = T, the ratio of their activities decreases according to e-3t. If the half-life of A is In 2, the half-life of B will be

  1. 4 In 2
  2. \(\frac{1}{4} \ln 2\)
  3. \(\frac{1}{2} \ln 2\)
  4. 2 In 2

Answer: 2. \(\frac{1}{4} \ln 2\)

For the given samples of A and B,

⇒ \(A_{\mathrm{A}}=A_0 \mathrm{e}^{-\lambda_{\mathrm{A}} t} \text { and } A_{\mathrm{B}}=A_0 \mathrm{e}^{-\lambda_{\mathrm{B}} t}\)

∴ \(\frac{A_{\mathrm{B}}}{A_{\mathrm{A}}}=\mathrm{e}^{\left(\lambda_{\mathrm{A}}-\lambda_{\mathrm{B}}\right) t}\)

Given that \(\frac{A_{\mathrm{B}}}{A_{\mathrm{A}}}=\mathrm{e}^{-3 t}\)

Hence, \(\left(\lambda_{\mathrm{A}}-\lambda_{\mathrm{B}}\right) t=-3 t\)

But \(\lambda_{\mathrm{A}}=\frac{\ln 2}{T_{\mathrm{A}}}=\frac{\ln 2}{\ln 2}=1\)

So, \(\lambda_{\mathrm{B}}=4\)

∴ \(T_{\mathrm{B}}=\frac{\ln 2}{\lambda_{\mathrm{B}}}=\frac{1}{4} \ln 2\).

Question 31. The (9/16)th part of a radioactive sample remains active after a time t. What fraction of the same remains undecayed after a time t/2?

  1. \(\frac{9}{16}\)
  2. \(\frac{5}{8}\)
  3. \(\frac{3}{4}\)
  4. \(\frac{4}{7}\)

Answer: 3. \(\frac{3}{4}\)

In the first case,

⇒ \(\frac{9}{16} N_0=N_0 \mathrm{e}^{-\lambda t} \Rightarrow \frac{9}{16}=\mathrm{e}^{-\lambda t}\)

In the second case,

⇒ \(x N_0=N_0 \mathrm{e}^{-\lambda t / 2} \Rightarrow x^2=\mathrm{e}^{-\lambda t}\)

∴ \(x^2=\frac{9}{16} \text { and } x=\frac{3}{4}\)

Question 32. A radioactive nucleus A decays into B (having a half-life of 10 s) and into C (having a half-life of 100 s). The equivalent half-life for both emissions is

  1. 6 s
  2. 9 s
  3. 2 s
  4. 3 s

Answer: 2. 9 s

The rate of disintegration is given by

⇒ \(-\frac{d N}{d t}=\lambda_1 N+\lambda_2 N=\left(\lambda_1+\lambda_2\right) N\)

∴ \(\lambda_{\text {eq }} N=\left(\lambda_1+\lambda_2\right) N\)

⇒ \(\lambda_{\text {eq }}=\lambda_1+\lambda_2 \Rightarrow \frac{\ln 2}{T}=\frac{\ln 2}{T_1}+\frac{\ln 2}{T_2}\)

∴ \(T=\frac{T_1 T_2}{T_1+T_2}=\frac{(10 \mathrm{~s})(100 \mathrm{~s})}{110 \mathrm{~s}}=\frac{100}{11} \mathrm{~s} \approx 9 \mathrm{~s}\).

33. The activities of three radioactive samples A, B, and C are represented by the lines A, B, and C in the figure. Their half-lives \(T_{1 / 2}(\mathrm{~A}), T_{1 / 2}(\mathrm{~B}), T_{1 / 2}(\mathrm{C}) \) are in the ratio

Radioactivity Multiple Choice Questions And Answers Three Radioactive Samples Q33

  1. 2: 1 : 3
  2. 4 : 3: 1
  3. 3 : 2: 1
  4. 2: 1: 1

Answer: 1. 2: 1 : 3

Activity = \(|R|=R_0 \mathrm{e}^{-\lambda t}\)

∴ \(\ln R=\ln R_0-\lambda t=\ln R_0-\left(\frac{\ln 2}{T}\right) t\)

The slope of the given graph is -X.

∴ \(\lambda_{\mathrm{A}}=\frac{6}{10}, \lambda_{\mathrm{B}}=\frac{6}{5} \text { and } \lambda_{\mathrm{C}}=\frac{2}{5}\)

∴ \(T_{\mathrm{A}}: T_{\mathrm{B}}: T_{\mathrm{C}}=\frac{10}{6}: \frac{5}{6}: \frac{5}{2}=2: 1: 3\)

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