If 10% of a radioactive material decays in 5, | vedclass

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Question

$10 \%$ decays in $5$ days $\Rightarrow \frac{\mathrm{N}^{1}}{\mathrm{N}_{0}}=\frac{10}{100}$

Thus active fraction in $5$ days $=\frac{\mathrm{N}}{

Vedclass · Accepted Answer

$65$

Vedclass · Answer

$10 \%$ decays in $5$ days $\Rightarrow \frac{\mathrm{N}^{1}}{\mathrm{N}_{0}}=\frac{10}{100}$

Thus active fraction in $5$ days $=\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}ight)^{5 / T_{\mathrm{H}}}$

$\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{90}{100}=\left(\frac{1}{2}ight)^{5 / T_{\mathrm{H}}}$

Thus amount of left material after $20$ days

$\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}ight)^{20 / \mathrm{T}_{\mathrm{H}}}=\left(\frac{1}{2}ight)^{5 / T_{\mathrm{H}} 	imes 4}$

$=\left(\frac{9 
ot 0}{10 
ot 0}ight)^{4}=\left(\frac{9}{10}ight)^{4}=0.6561$

in percentage approx $=65.61 \%$

If $10\%$ of a radioactive material decays in $5\, days$ then the amount of the original material left after $20\, days$ is approximately .......... $\%$

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