Question

In a radioactive substance at $t=0$, the number of atoms is $8\times {10}^4$. Its half-life period is $3$ years. The number of atoms $1\times {10}^4$ will remain after interval.

• A. $9$ years
• B. $8$ years
• C. $6$ years
• D. $24$ years

Answer 1

Answer: (A)

$9$ years

$N=N_0{\left(\frac{1}{2}\right)}^{t/\tau }$

${10}^4=8\times {10}^4{\left(\frac{1}{2}\right)}^{t/3}$

$\left(\frac{1}{8}\right)={\left(\frac{1}{2}\right)}^{t/3}$

${\left(\frac{1}{2}\right)}^3={\left(\frac{1}{2}\right)}^{t/3}=3=\frac{t}{3}$

t is 9 yr