The half life of a radioactive isotope 'X | vedclass

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Question

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Vedclass · Accepted Answer

$60$

Vedclass · Answer

$X$        $	o $           $Y$

Initial number of atoms,                            $N_0$                     $N$

Number of atoms after time $t$                $N$                         $N_0-N$

As per question

$\frac{N}{{{N_0} - N}} = \frac{1}{7} \Rightarrow 7N = {N_0} - N$  or  $8 N=N_{0}$

${\frac{N}{N_{0}}=\frac{1}{8}}$

As $\frac{N}{N_{0}}=\left(\frac{1}{2}ight)^{n}$ where $n$ is the no. of half lives

$	herefore \quad \frac{1}{8} = {\left( {\frac{1}{2}} ight)^n}$ or ${\left( {\frac{1}{2}} ight)^3} = {\left( {\frac{1}{2}} ight)^n}$

$	herefore n=3$

$n=\frac{t}{T_{1 / 2}} 	ext { or } t=n T_{1 / 2}=3 	imes 20 	ext { years }=60 	ext { years }$

Hence, the age of rock is $60$ years.

The half life of a radioactive isotope $'X'$ is $20$ years, It decays to another element $'Y'$ which is stable. The two elements $'X'$ and $'Y'$ were found to be in the ratio $1:7$ in a simple of a given rock . The age of the rock is estimated to be............$years$

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