Question 421: Atoms & Nuclei

Question

The activity of radioactive sample is measured as $\mathrm{N}_0$ counts per minute at time $\mathrm{t}=0$, and $\frac{\mathrm{N}_0}{\mathrm{e}}$ counts per minute at time $\mathrm{t}=3$ minute, The activity reduces to half its value in time (in minute)

Accepted Answer

Step 1: Write the decay law
The activity at time $t$ is given by:
$ N(t) = N_0 e^{-\lambda t} $
where $\lambda$ is the decay constant.
Step 2: Find decay constant from given data
At $t=3$ min,
$ N(3) = \frac{N_0}{e} $
So,
$ \frac{N_0}{e} = N_0 e^{-3\lambda} $
Cancel $N_0$:
$ \frac{1}{e} = e^{-3\lambda} $
$ e^{-1} = e^{-3\lambda} $
$ -1 = -3\lambda \quad \Rightarrow \quad \lambda = 	frac{1}{3} $
So, the decay constant is:
$ \lambda = \frac{1}{3}\,	ext{per minute} $
Step 3: Find half-life
The half-life is:
$ T_{1/2} = \frac{\ln 2}{\lambda} $
Substitute $\lambda = \frac{1}{3}$:
$ T_{1/2} = \frac{\ln 2}{1/3} = 3 \ln 2 $
Final Answer:
$ \boxed{3 \ln 2} $
That corresponds to
Option B
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Answer

$\frac{1}{3} \log _{\mathrm{e}} 2$

Answer

$3 \log _{\mathrm{e}} 2$

Answer

$3 \log _{10} 2$

Answer

$\frac{3}{\log _{10} 2}$

Atoms & Nuclei

The activity of radioactive sample is measured as \mathrm{N}_0 counts per minute at time \mathrm{t}=0, and \frac{\mathrm{N}_0}{\mathrm{e}} counts per minute at time \mathrm{t}=3 minute, The activity reduces to half its value in time (in minute)

Choose the correct answer: