Question

If 10% of a radioactive material decays in 5 days, then the amount of the original material left after 20 days is approximately:

• A. 60%
• B. 65%
• C. 70%
• D. 75%

Answer 1

Answer: (B)

The correct option is B 65%
Let the initial amount of radioactive substance be $N_o$.
Using $\frac{0.693}{\lambda }t=\ln \frac{N_o}{N}$
After $t=5$ days, $N=0.9N_o$
$\therefore$ $\frac{0.693}{\lambda }\times 5=\ln \frac{N_o}{0.9N_o}$ .......(1)
Using $\frac{0.693}{\lambda }t=\ln \frac{N_o}{N}$
After $t=20$ days :
$\therefore$ $\frac{0.693}{\lambda }\times 20=\ln \frac{N_o}{N}$ ........(2)
Dividing equation (1) and (2), we get $\frac{5}{20}=\frac{\ln \frac{N}{0.9N_o}}{\ln \frac{N}{N_o}}$
Or $\frac{1}{4}=\frac{0.104}{\ln \frac{N_o}{N}}$
Or $\ln \frac{N_o}{N}=0.104\times 4=0.416$
We get $N=\frac{N_o}{e^{0.416}}=0.65$
Thus $65$% of radioactive substance decays after 20 days.