Question

To determine the half life of a radioactive element, a student plots a graph of $ln|\frac{dN\left(t\right)}{dt}|$ versus t. Here $\frac{dN\left(t\right)}{dt}$ is the rate of radioactive decay at time t. If the number of radioactive nuclei of this element decreases by a factor of x after 4.16 years, the value of x is

• A. 4
• B. 6
• C. 8
• D. 10

Answer 1

Answer: (C)

8

$\begin{array}{c}\frac{dN}{dt}=-\lambda N⟶\left(i\right)\\ \text{also,}\\ \begin{array}{c}N=N,e^{-\lambda t}⟶\left(ii\right)\\ \lambda N=\lambda N_0{e}^{-\lambda t}\\ \ln \left(\lambda N\right)=\ln \lambda N_0-\lambda t\end{array}\end{array}$

$\begin{array}{c}\ln \left(\frac{dN}{dt}\right)=\ln \lambda N_0-\lambda t⟶\left(iii\right)\\ \text{slope}=-\lambda \\ \text{from graph, slope}=\frac{4-3}{4-6}=\frac{-1}{2}\\ \begin{array}{c}\text{after 4.16 years}\end{array}\\ N=N_0{e}^{-\frac{1}{2}\times 4\cdot 16}\end{array}$

$\frac{N}{N_0}=\frac{1}{e^{2.08}}=\frac{1}{8}$

$\begin{array}{c}\text{Hence, the number of radioactive nuclei decreases by}\text{a factor of}8\end{array}$