Question

There are n number of radioactive nuclei in a sample at a time t. If n' number of $\beta$ particles are emitted in the next two seconds from the sample, then the half life of the sample is ($t_{half \ life}>>2$ sec):

• A. $\frac{n'}{2}$
• B. $0.693 \times (\frac{2n}{n'})$
• C. $0.693 \times ln(\frac{2n}{n'})$
• D. $0.693 \times (\frac{n}{n'})$

Answer 1

Answer: (B)

Option B

Answer 2

Given that the number of decays in 2 seconds is $n'$ and there are $n$ nuclei at time $t$, the activity is nearly constant because the half‐life is much larger than 2 seconds. Thus, we have:

$$n' = \lambda\, n \times 2 \quad \Rightarrow \quad \lambda = \frac{n'}{2n}$$

The half-life $t_{1/2}$ is related to the decay constant $\lambda$ by:

$$t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\frac{n'}{2n}} = 0.693 \times \left(\frac{2n}{n'}\right)$$