First slide

Radio activity

Question

The time elapsed between 25% and 50% decay of a radioactive sample having half life of 1 hour is

Difficult

A

$(\frac{\ln 3 - \ln 2}{\ln 2}) h r$
B

$(\frac{\ln 3}{\ln 2}) h r$
C

$\frac{2 \ln 2}{\ln 3} h r$
D

$(\ln 4 - \ln 3) h r$

Solution

$N_{1} = N_{0} - 25 % of N_{0} = \frac{3 N_{0}}{4}$
and $N_{2} = N_{0} - 50 % of N_{0} = \frac{N_{0}}{2}$
$∴ \frac{3 N_{0}}{4} = N_{0} e^{- {λt}_{1}} \Rightarrow e^{{λt}_{1}} = \frac{4}{3} \Rightarrow t_{1} = \frac{\ln \frac{4}{3}}{λ} = T . (\frac{\ln \frac{4}{3}}{\ln 2})$
Again, $\frac{N_{0}}{2} = N_{0} e^{- {λt}_{2}} \Rightarrow e^{{λt}_{2}} = 2 \Rightarrow t_{2} = \frac{\ln 2}{λ} = T$
$∴ t_{2} - t_{1} = T - T . \frac{\ln \frac{4}{3}}{\ln 2} = T \frac{\ln 2 - 2 \ln 2 + \ln 3}{\ln 2}$
$\Rightarrow t_{2} - t_{1} = 1 . \frac{\ln 3 - \ln 2}{\ln 2} hr$

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