Half life of a radio-active substance is 20, | vedclass

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Question

(b) Here ${T_{1/2}} = 20\,minutes$;

we know $\frac{N}{{{N_0}}} = {\left( {\frac{1}{2}} \right)^{t/{T_{1/2}}}}$

For $20\%$ decay $\frac{N}{{{N_0}

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(b) Here ${T_{1/2}} = 20\,minutes$;

we know $\frac{N}{{{N_0}}} = {\left( {\frac{1}{2}} \right)^{t/{T_{1/2}}}}$

For $20\%$ decay $\frac{N}{{{N_0}}} = \frac{{80}}{{100}} = {\left( {\frac{1}{2}} \right)^{{t_1}/20}}$..... $(i)$

For $80\%$ decay $\frac{N}{{{N_0}}} = \frac{{20}}{{100}} = {\left( {\frac{1}{2}} \right)^{{t_2}/20}}$..... $(ii)$

Dividing $ (ii)$ by $(i)$

$\frac{1}{4} = {\left( {\frac{1}{2}} \right)^{\frac{{({t_2} - {t_1})}}{{20}}}};$

on solving we get ${t_2} - {t_1} = 40\,min.$

Half life of a radio-active substance is $20\, minutes$. The time between $20\%$ and $80\%$ decay will be ........... $minutes$

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