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 $t=2hr,$
$N_1=8\times 10^{10}{\left[\frac{1}{2}\right]}^{\frac{2}{1}}$
$=2\times 10^{10}$
No. of atoms at $t=4hr$
$N_2=8\times 10^{10}{\left[\frac{1}{2}\right]}^{\frac{4}{1}}$
$=\frac{1}{2}\times 10^{10}$
$\therefore$ No. of atoms decayed in given duration
$=\left(2-\frac{1}{2}\right)\times 10^{10}=1.5\times 10^{10}$