Answer to Question #118724 in Atomic and Nuclear Physics for Neha

Question

You have a radioactive sample which has an activity of 3.89×105 Bq at t = 0 s. Later on at t = 1540 s you measure the activity again and find it is 3.15×105 Bq.

What is the half-life of the sample?

Answer 1

Given, initial activity i.e activity of the radio active element at $t=0s$ is $A=A_0=3.89\times 10^5Bq$ , and at time $t=1540s$ ,activity $A=3.15\times10^5Bq$ .

Since, Activity of radio active elements is governed by at any time $t$ is given by,

$$A=A_0e^{-\lambda t}$$

where,$\lambda$ is decay constant.

Thus, we get,

$$3.15\times10^5=3.89\times10^5e^{-1540\lambda}\\ \implies\frac{3.15}{3.89}=e^{-1540\lambda}\\ \implies \lambda=\frac{1}{1540}ln\bigg(\frac{3.89}{3.15}\bigg)\\ \implies \lambda=1.37\times 10^{-4}s^{-1}$$

Now, we also know that, half life of radio active element is given by,

$$t_{\frac{1}{2}}=\frac{ln2}{\lambda}\\ \implies t_{\frac{1}{2}}=\frac{0.963}{1.37\times10^{-4}}=5059.4\approx5059s$$

Thus,

$$\boxed{t_{\frac{1}{2}}=5059s}$$