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. $2 \times 10^{10}$
• B. $1.5 \times 10^{10}$
• C. Zero
• D. Infinity

Answer 1

Answer: (B)

$1.5 \times 10^{10}$

Answer 2

$N = N_{0}\left( \frac{1}{2} \right)^{\frac{t}{T_{1l2}}}$

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, $N_{2} = 8 \times 10^{10}\left( \frac{1}{2} \right)^{\frac{4}{1}} = \frac{1}{2} \times 10^{10}$

∴ No. of atoms decayed in given duration

$= \left( 2 - \frac{1}{2} \right) \times 10^{10} = 1.5 \times 10^{10}$