First slide

Radio activity

Question

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

Moderate

A

$2 x 10^{10}$
B

$1.5 x 10^{10}$
C

Zero
D

Infinity

Solution

$N = N_{0} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$
No. of atoms at t=2hr,
$N_{1} = 8 x 10^{10} {(\frac{1}{2})}^{\frac{2}{1}} = 2 x 10^{10}$
No. of atoms at t=4hr,
$N_{2} = 8 x 10^{10} {(\frac{1}{2})}^{\frac{4}{1}} = \frac{1}{2} x 10^{10}$
$∴$ No. of atoms decayed in given duration
$= (2 - \frac{1}{2}) x 10^{10} = 1.5 x 10^{10}$

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