Question

In a radioactive sample the fraction of initial concentration which remains undecayed after 'n' mean lives equals:

• A. ${\left(1/e\right)}^n$
• B. ${(1-1/e)}^n$
• C. $1-\frac{1}{e^n}n$
• D. ${\left(\frac{1}{e-1}\right)}^n$

Answer 1

Answer: (A)

${\left(1/e\right)}^n$

$\begin{array}{c}N\to \text{Final concentration}\\ N_0\to \text{Initial concentration}\\ \lambda \to \text{decay constant}\\ t\to \text{time}\\ z\to \text{mean life}\end{array}$

$\begin{array}{c}\text{According to equation of radioactivity}\\ \begin{array}{c}N=N_0{e}^{-\lambda t}⟶\left(i\right)\\ \tau =\frac{1}{\lambda }\\ n\tau =\frac{n}{\lambda }\\ \text{from (i)}\\ \frac{N}{N_0}=e^{\lambda \times \frac{n}{\lambda }}\\ \frac{N!}{N_0}=e^{-n}=\frac{1}{e^n}\end{array}\end{array}$