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Question

Given: $\frac{\mathrm{d} \mathrm{N}_{0}}{\mathrm{dt}}=20$ decays $/ \mathrm{mir}$

$\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}=2$ decays/min

$\mat

Vedclass · Accepted Answer

$19039$

Vedclass · Answer

Given: $\frac{\mathrm{d} \mathrm{N}_{0}}{\mathrm{dt}}=20$ decays $/ \mathrm{mir}$

$\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}=2$ decays/min

$\mathrm{T}_{1 / 2}=5730 \mathrm{years}$

As we know,

$\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t}$

$\log \frac{N_{0}}{N}=\lambda t$

$	herefore \mathrm{t}=\frac{1}{\lambda} \log \frac{\mathrm{N}_{0}}{\mathrm{N}}$

$ = \frac{{2.303 	imes {{	ext{T}}_{1/2}}}}{{0.693}} 	imes {	ext{Lo}}{{	ext{g}}_{10}}\frac{{{{	ext{N}}_0}}}{{	ext{N}}}$

But $\frac{\frac{d N_{0}}{d t}}{\frac{d N}{d t}}=\frac{N_{0}}{N}=\frac{20}{2}=10$

$	herefore \,\,t = \frac{{2.303 	imes 5730}}{{0.693}} 	imes 1$

$=19039$ years

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