રેડિયોએક્ટિવ નમૂનામાં શરૂઆતમાં 10^{20} અણુઓ છ | vedclass

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Question

$t = 0 $              $N_0 = 10^{20}$

$t = 1y$             $  N_1 = N_0e

Vedclass · Accepted Answer

$7	imes 10^{19}$

Vedclass · Answer

$t = 0 $              $N_0 = 10^{20}$

$t = 1y$             $  N_1 = N_0e^{-\lambda(1)}$

$t = 2y$             $  N_2 =N_0e^{-\lambda(2)}$

$t = 3y$            $   N_3 = N_0e^{\lambda(3)}$

$\frac{{{N_2} - {N_3}}}{{{N_1} - {N_2}}} = 0.3\,\,\,$

$ \Rightarrow  \frac{{{N_0}{e^{ - 2\lambda }} - {N_0}{e^{ - 3\lambda }}}}{{{N_0}{e^{ - \lambda }} - {N_0}{e^{ - 2\lambda }}}} = 0.3$

$\frac{{{N_0}{e^{ - 2\lambda }}(1 - {e^{ - \lambda }})}}{{{N_0}{e^{ - \lambda }}(1 - {e^{ - \lambda }})}} = 0.3$

$e^{-\lambda}  = 0.3   $

$	herefore N_0 - N_1  = ?$

$= N_0 - N_0e^{-\lambda} = N_0(1 - e^{-\lambda})$  

$=  10^{20} (1 - 0.3) = 0.7 	imes 10^{20} = 7 	imes 10^{19}$

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