Question

If $20\%$ of a radioactive substance decays in $5$ days, then find the approximate percentage of the initial amount left after $20$ days.

• A. $41\%$
• B. $82\%$
• C. $39\%$
• D. $59\%$

Answer 1

The correct option is A $41\%$

$\because \ln \frac{A_o}{A_t}=-kt\ln \left(\frac{100}{80}\right)=-k\times 5-k=\frac{1}{5}\ln \frac{5}{4}k=\frac{1}{5}\ln \frac{4}{5}So,\ln \left(\frac{100}{x}\right)=-\frac{1}{5}\ln \frac{4}{5}\times 20\ln \left(\frac{x}{100}\right)=4\ln \frac{4}{5}\ln \left(\frac{x}{100}\right)=\ln {\left(\frac{4}{5}\right)}^4\frac{x}{100}=\frac{4\times 4\times 4\times 4}{5\times 5\times 5\times 5}\Rightarrow x=40.96$