Question

A freshly prepared sample of a radioisotope of half-life $1386s$ has activity ${10}^{3}disintegrationspersecond.$ Given that $\ell n2=0.693,$ the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will decay in the first $80s$ after preparation of the sample is:

• A. $1$%
• B. $2$%
• C. $3$%
• D. $4$%

Answer 1

The correct option is D $4$%

$\begin{array}{c}\text{Let,Decay constant be}\lambda .\\ \lambda =\frac{\ln 2}{T/2}=\frac{0.693}{1386}\\ \text{Initial Activity}={10}^3\\ \therefore \frac{0.693}{1386}{N}_0={10}^3\\ N=N_0{e}^{-\lambda t}\end{array}$

$\text{After 80 sec,}$

$\begin{array}{cc}\frac{N}{N_0}& =e^{-80\lambda }\\ \%\frac{N}{N_0}& =\frac{100}{e^{80\lambda }}=3.921\%\\ & \simeq 4\%.\end{array}$