Radioactive Decay Laws Questions and Answers

Q: A radioactive isotope sample has a decay constant \(\lambda = 2.31 \times 10^{-4}\) year-1. Initially the sample contains \(8000 \times 10^{15}\) atoms. The sample decays to 25% of its initial value. The number of years it takes to decay to this stage is closest to ___

The time is closest to 6000 years.Radioactive decay follows \(N = N_{0}e^{-\lambda t}\). Decaying to 25% means \(N/N_{0} = 1/4\):\[ \begin{aligned} t &= \frac{\ln 4}{\lambda} = \frac{1.386}{2.31 \times 10^{-4}} \\ &\approx 6000\ \text{years} \end{aligned} \]Equivalently, the half-life is \(0.693/\lambda = 3000\) years, and falling to a quarter takes two half-lives. 3000 is one half-life, the time to 50%. The initial number of atoms does not matter.

Q: The radioactive iodine (131I) has a half life of eight days. The amount of 0.5 g sample left after 32 days would be :

The amount left is 31.25 mg.32 days is four half-lives of 8 days, and each half-life halves the amount:\[ \begin{aligned} \text{left} &= 0.5 \text{ g} \times \left(\tfrac{1}{2}\right)^{4} \\ &= 0.03125 \text{ g} = 31.25 \text{ mg} \end{aligned} \]0.25 g remains after one half-life, 125 mg after two and 62.5 mg after three; those options stop counting too early.

Q: A radioactive source with half-life \((\tau_{1/2})\) of 69.3 sec has an activity of \(1\mu Ci\). How many radioactive nuclei will remain after time t = 69.3 sec ?

The number remaining is \(1.85 \times 10^{6}\).The activity fixes the initial number of nuclei through \(A = \lambda N\), with \(\lambda = \ln 2/\tau_{1/2}\):\[ \begin{aligned} \lambda &= \frac{0.693}{69.3\ \text{s}} = 0.01\ \text{s}^{-1} \\ N_{0} &= \frac{A}{\lambda} = \frac{3.7 \times 10^{4}\ \text{s}^{-1}}{0.01\ \text{s}^{-1}} \\ &= 3.7 \times 10^{6} \end{aligned} \]using \(1\ \mu Ci = 3.7 \times 10^{4}\) decays per second. After one half-life, 69.3 s, half remain: \(1.85 \times 10^{6}\).\(3.7 \times 10^{6}\) is the initial number, before any decay. The options with \(10^{5}\) slip a power of ten in the curie conversion.

A radioactive isotope sample has a decay constant \(\lambda = 2.31 \times 10^{-4}\) year^-1. Initially the sample contains \(8000 \times 10^{15}\) atoms. The sample decays to 25% of its initial value. The number of years it takes to decay to this stage is closest to ___

A.3000
B.6000
C.900
D.2600

Exam Year(s)

GATE XE - 2026,

Answer: B

The time is closest to 6000 years.

Radioactive decay follows \(N = N_{0}e^{-\lambda t}\). Decaying to 25% means \(N/N_{0} = 1/4\):

\[ \begin{aligned} t &= \frac{\ln 4}{\lambda} = \frac{1.386}{2.31 \times 10^{-4}} \\ &\approx 6000\ \text{years} \end{aligned} \]

Equivalently, the half-life is \(0.693/\lambda = 3000\) years, and falling to a quarter takes two half-lives. 3000 is one half-life, the time to 50%. The initial number of atoms does not matter.

Discuss in Forum

The radioactive iodine (¹³¹I) has a half life of eight days. The amount of 0.5 g sample left after 32 days would be :

A.125 mg
B.62.5 mg
C.31.25 mg
D.0.25 g

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The amount left is 31.25 mg.

32 days is four half-lives of 8 days, and each half-life halves the amount:

\[ \begin{aligned} \text{left} &= 0.5 \text{ g} \times \left(\tfrac{1}{2}\right)^{4} \\ &= 0.03125 \text{ g} = 31.25 \text{ mg} \end{aligned} \]

0.25 g remains after one half-life, 125 mg after two and 62.5 mg after three; those options stop counting too early.

Discuss in Forum

A radioactive source with half-life \((\tau_{1/2})\) of 69.3 sec has an activity of \(1\mu Ci\). How many radioactive nuclei will remain after time t = 69.3 sec ?

A.\(3.7 \times 10^{6}\)
B.\(3.7 \times 10^{5}\)
C.\(1.85 \times 10^{6}\)
D.\(1.85 \times 10^{5}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

The number remaining is \(1.85 \times 10^{6}\).

The activity fixes the initial number of nuclei through \(A = \lambda N\), with \(\lambda = \ln 2/\tau_{1/2}\):

\[ \begin{aligned} \lambda &= \frac{0.693}{69.3\ \text{s}} = 0.01\ \text{s}^{-1} \\ N_{0} &= \frac{A}{\lambda} = \frac{3.7 \times 10^{4}\ \text{s}^{-1}}{0.01\ \text{s}^{-1}} \\ &= 3.7 \times 10^{6} \end{aligned} \]

using \(1\ \mu Ci = 3.7 \times 10^{4}\) decays per second. After one half-life, 69.3 s, half remain: \(1.85 \times 10^{6}\).

\(3.7 \times 10^{6}\) is the initial number, before any decay. The options with \(10^{5}\) slip a power of ten in the curie conversion.

Discuss in Forum

Loading…

Topics

Other Subjects

Tags: Radioactive Decay Laws Questions and Answers || Radioactive Decay Laws MCQ Questions and Answers || Radioactive Decay Laws GK Questions and Answers || Radioactive Decay Laws GK MCQ Questions || Radioactive Decay Laws Multiple Choice Questions and Answers || Radioactive Decay Laws GK || GK on Radioactive Decay Laws || Nuclear Physics Questions and Answers || Nuclear Physics MCQ Questions and Answers || Nuclear Physics GK Questions and Answers || GK on Nuclear Physics

Radioactive Decay Laws Questions and Answers

Topics

Other Subjects

About Us

Stay informed about all important Dates & Notifications