Question

The half life of a sample of a radioactive substance is 1 hour. If $8 \times 10^{10}$ atoms are present at $t=0$, then the number of atoms decayed in the duration $t =2$ hour to $t =4$ hour will be:

• A. $1.7 \times 10^{7}$
• B. $1.5 \times 10^{10}$
• C. $3 \times 10^{7}$
• D. $2.8 \times 10^{10}$

Answer 1

Answer: (B)

$ N = N _{0}\left(\frac{1}{2}\right)^{\frac{ t }{ t } / 2}$
No. of atoms at $t=2 h r, $
$N_{1}=8 \times 10^{10}\left[\frac{1}{2}\right]^{\frac{2}{1}}$
$=2 \times 10^{10}$
No. of atoms at $t=4 h r$
$N _{2}=8 \times 10^{10}\left[\frac{1}{2}\right]^{\frac{4}{1}}$
$=\frac{1}{2} \times 10^{10}$
$\therefore $ No. of atoms decayed in given duration
$=\left(2-\frac{1}{2}\right) \times 10^{10}=1.5 \times 10^{10}$