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13.Nuclei
hard
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
Solution
(b) $N = {N_0}{\left( {\frac{1}{2}} \right)^{\frac{t}{{{T_{1l2}}}}}}$
No of atoms at $t = 2\,hr$,
${N_1} = 8 \times {10^{10}}{\left( {\frac{1}{2}} \right)^{\frac{2}{1}}} = 2 \times {10^{10}}$
No. of atoms at $t = 4\,hr$,
${N_2} = 8 \times {10^{10}}{\left( {\frac{1}{2}} \right)^{\frac{4}{1}}} = \frac{1}{2} \times {10^{10}}$
No. of atoms decayed in given duration
$ = \left( {2 – \frac{1}{2}} \right) \times {10^{10}} = 1.5 \times {10^{10}}$
Standard 12
Physics
