Question

A radioactive substance decays 20% in 10 min if at start there are $5\times {10}^{20}$ atoms present, after what time will the number of atoms be reduced to ${10}^{18}$ atoms?

• A. 4.65 hour
• B. 2.32 hour
• C. 9.3 hour
• D. 2 hour

Answer 1

The correct option is A 4.65 hour

In 10 minutes, 20% of radioactive substance decays. So 80% of the radioactive substance remains.
$a=5\times {10}^{20}$ atoms
$a-x=5\times {10}^{20}\times \frac{80}{100}$ atoms
The integrated rate law expression for the first order decay process is
$k=\frac{2.303}{t}log\frac{a}{a-x}$
$k=\frac{2.303}{10min}log\frac{5\times {10}^{20}}{5\times {10}^{20}\times \frac{80}{100}}$
$k=0.022318/min$
After time t, the number of atoms will be reduced to ${10}^{18}$ atoms
$t=\frac{2.303}{k}log\frac{a}{a-x}$
$t=\frac{2.303}{0.022318/min}log\frac{5\times {10}^{20}}{{10}^{18}}$
$t=278.5min$
Convert the unit of time from min to hours
$t=\frac{278.5min}{60min/hr}=4.65hour$