Question

A radioactive element decays by beta emission. A detector records n beta particle in the first 2 seconds and 0.75 n beta particles in the next 2 seconds. The decay constant of the element is

• A. $\lambda =\frac{1}{2}In\left(\frac{4}{3}\right)$
• B. $\lambda =\frac{1}{2}In\left(\frac{3}{4}\right)$
• C. $\lambda =\frac{1}{2}In\left(2\right)$
• D. $\lambda =\frac{3}{4}In\left(2\right)$

Answer 1

The correct option is A

$\lambda =\frac{1}{2}In\left(\frac{4}{3}\right)$

$N_t=N_0{e}^{-\lambda t}$ Therefore

$N_2=N_0{e}^{-2\lambda }and{N}_4=N_0{e}^{-4\lambda }$

Given $n=N_0-N_2=N_0(1-e^{-2\lambda })\left(1\right)$

and $0.75n=N_2-N_4=N_0{e}^{-2\lambda }-N_0{e}^{-4\lambda }$

$\Rightarrow 0.75n=N_0(e^{-2\lambda }-e^{-4\lambda })$

$\Rightarrow 0.75n=N_0{e}^{-2\lambda }(1-e^{-2\lambda })\left(2\right)$

From Eqs. (1) and (2) we get

$0.75=e^{-2\lambda }\Rightarrow e^{2\lambda }=\frac{4}{3}\Rightarrow \lambda =\frac{1}{2}ln\left(\frac{4}{3}\right)$