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13.Nuclei
hard
The half life of a radioactive substance is $20$ minutes. The approximate time interval $(t_2 -t_1)$ between the time $t_2$ when $3/4$ of it has decayed and time $t_1$ when $1/4$ of it had decayed is
A
$\frac{{20}}{{\ln 2}}\min$
B
$\frac{{20\ln 3}}{{\ln 2}}\min$
C
$20$ $min$
D
$20\ln 2\min$
Solution
$\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$
$\frac{3}{4}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t_{2}}$ …….$(1)$
$\frac{1}{4}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t_{1}}$ ……..$(2)$
$\lambda=\frac{\ln 2}{20}$ ………..$(3)$
Solve equation $(1),(2) $ and $(3)$
$t_{2}-t_{1}=\frac{20 \ln 3}{\ln 2}$
Standard 12
Physics
