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(c) ${\lambda _\alpha } = \frac{1}{{1620}}$ per year and ${\lambda _\beta } = \frac{1}{{405}}$ per year and it is given that the fraction of the remai

Vedclass · Accepted Answer

$449$

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(c) ${\lambda _\alpha } = \frac{1}{{1620}}$ per year and ${\lambda _\beta } = \frac{1}{{405}}$ per year and it is given that the fraction of the remained activity $\frac{A}{{{A_0}}} = \frac{1}{4}$

Total decay constant

$\lambda = {\lambda _\alpha } + {\lambda _\beta } = \frac{1}{{1620}} + \frac{1}{{405}} = \frac{1}{{324}}per\,year$

We know that $A = {A_0}{e^{ - \lambda t}}$==> $t = \frac{1}{\lambda }{\log _e}\frac{{{A_0}}}{A}$

==>$t = \frac{1}{\lambda }{\log _e}4 = \frac{2}{\lambda }{\log _e}2$

$=324 &times; 2 &times; 0.693 = 449\, years.$

For a substance the average life for $\alpha$-emission is $1620$ years and for $\beta$ emission is $405$ years. After how much time the $1/4$ of the material remains after $\alpha$ and $\beta$ emission .......$years$

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