Chapter 6.4, Problem 50E

To determine

To choose : The correct option.

Answer to Problem 50E

The correct option is $(C)$ .

Explanation of Solution

Given information:

The sample of cerium- $143$ losses $99\%$ of its radioactive matter in $199$ hours .

Calculation:

If the half-life is $t_{1/2}$ , then

  $\frac{N}{N_0}={\left(\frac{1}{2}\right)}^{t/t_{1/2}}$

Put, $N=0.01N_0$

Then,

  $\begin{array}{c}\frac{0.01N_0}{N_0}={\left(\frac{1}{2}\right)}^{199/t_{1/2}}\\ \Rightarrow 0.01={\left(\frac{1}{2}\right)}^{199/t_{1/2}}\\ \Rightarrow \ln 0.01=\ln {\left(\frac{1}{2}\right)}^{199/t_{1/2}}\\ \Rightarrow \ln 0.01=\frac{199}{t_{1/2}}\cdot \ln \left(\frac{1}{2}\right)\\ \Rightarrow \ln 0.01=\frac{-199\ln 2}{t_{1/2}}\\ \Rightarrow \frac{\ln 0.01}{-199\ln 2}=\frac{1}{t_{1/2}}\\ \Rightarrow -\frac{199\ln 2}{\ln 0.01}=t_{1/2}\\ \Rightarrow t_{1/2}\approx 29.95\\ \approx 30\end{array}$

Therefore, the required correct option is $(C)$ .