Question

Ninety percent of an radioactive sample is left over after a time interval $t$. The percentage of initial sample that will disintegrate in an interval $2t$ is?

• A. $38$%
• B. $81$%
• C. $9$%
• D. $62$%

Answer 1

The correct option is B $81$%

b'$81$%

Formula,

at t,

$N=N_{0}exp(-\lambda t)$

$\frac{N}{N_0}=exp(-\lambda t)=0.9$

$\ln \left(0.9\right)=-\lambda t$

$\lambda =-\frac{\ln \left(0.9\right)}{t}$

at 2t

$\frac{N}{N_0}=exp(-2\lambda t)=exp\left(2t\frac{\ln \left(0.9\right)}{t}\right)=0.81$

$=81$%

'