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13.Nuclei
hard
Half life of a radio-active substance is $20\, minutes$. The time between $20\%$ and $80\%$ decay will be ........... $minutes$
A
$20$
B
$40$
C
$30$
D
$25$
Solution
(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 ${t_2} – {t_1} = 40\,min.$
Standard 12
Physics
