In a radioactive substance at t = 0, the numb | vedclass

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Question

(a) By formula $N = {N_0}{\left( {\frac{1}{2}} ight)^{t/T}}$

or ${10^4} = 8 	imes {10^4}{\left( {\frac{1}{2}} ight)^{t/3}}$

or $\left( {\fr

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(a) By formula $N = {N_0}{\left( {\frac{1}{2}} ight)^{t/T}}$

or ${10^4} = 8 	imes {10^4}{\left( {\frac{1}{2}} ight)^{t/3}}$

or $\left( {\frac{1}{8}} ight) = {\left( {\frac{1}{2}} ight)^{t/3}}$

or ${\left( {\frac{1}{2}} ight)^3} = {\left( {\frac{1}{2}} ight)^{t/3}}\,\, $

$\Rightarrow 3 = \frac{t}{3}$

Hence $t = 9\, years.$

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 ...........$years$

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