In a radioactive substance at t = 0 , number of atoms is 8 × {10^4} . Its half life period is 3 yea

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13.Nuclei

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In a radioactive substance at $t = 0$, the number of atoms is $8 \times {10^4}$. Its half life period is $3$ years. The number of atoms $1 \times {10^4}$ will remain after interval ...........$years$

A

$9$

B

$8$

C

$6$

D

$24$

Solution

(a) By formula $N = {N_0}{\left( {\frac{1}{2}} \right)^{t/T}}$

or ${10^4} = 8 \times {10^4}{\left( {\frac{1}{2}} \right)^{t/3}}$

or $\left( {\frac{1}{8}} \right) = {\left( {\frac{1}{2}} \right)^{t/3}}$

or ${\left( {\frac{1}{2}} \right)^3} = {\left( {\frac{1}{2}} \right)^{t/3}}\,\, $

$\Rightarrow 3 = \frac{t}{3}$

Hence $t = 9\, years.$

Standard 12

Physics

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