Physics - Modern Physics Question with Solution | TestHub

Radioactive decay equation is

$N=N_0{e}^{-\lambda t}=N_0{e}^{-\ln \left(2\right)\frac{t}{T_{\frac{1}{2}}}}$

After decay of 99% of the initial sample only 1% will be left, and$\frac{N}{N_0}=1\%$. We then have

$\frac{N}{N_0}=\frac{1}{100}=e^{-\ln \left(2\right)\frac{t}{T_{\frac{1}{2}}}}$

If we take the natural logarithm, we have

$-\ln 100=-\frac{\ln 2\times t}{T_{\frac{1}{2}}}$

Which on solving for$\text{t}$yields

$\therefore$$t=\frac{\ln 100}{\ln 2}\times T_{\frac{1}{2}}$

$=\frac{\log 100}{\log 2}\times T_{\frac{1}{2}}$

$=\frac{2}{0.3010}\times 110$

$=731\min =12.2 h$