Question 414: Atoms & Nuclei

Question

A radio active element has rate of disintegration 8000 disintegrations per minute at a particular instant. After four minutes it becomes 2000 disintegrations per minute. The decay constant per minute is

Accepted Answer

Given:
Initial rate of disintegration: $R_0 = 8000$ disintegrations/minute
Rate after $t = 4$ minutes: $R = 2000$
Decay law for activity $R$:
$ R = R_0 e^{-\lambda t} $
So,
$ \frac{R}{R_0} = e^{-\lambda t} $
$ \frac{2000}{8000} = e^{- \lambda (4)} $
$ \frac{1}{4} = e^{-4\lambda} $
Take logarithm:
$ -4\lambda = \ln\left(	frac{1}{4}ight) = -\ln 4 $
$ \lambda = \frac{\ln 4}{4} $
Now,
$ \ln 4 = \ln (2^2) = 2 \ln 2 $
So,
$ \lambda = \frac{2 \ln 2}{4} = \frac{1}{2} \ln 2 = 0.5 \ln 2 $
Ã¢ÂÂ Correct Option:
C. $0.5 \log_e 2$

Answer

$0.8 \log _{\mathrm{e}} 2$

Answer

$0.6 \log _{\mathrm{e}} 2$

Answer

$0.5 \log _{\mathrm{e}} 2$

Answer

$0.2 \log _{\mathrm{e}} 2$

Atoms & Nuclei

A radio active element has rate of disintegration 8000 disintegrations per minute at a particular instant. After four minutes it becomes 2000 disintegrations per minute. The decay constant per minute is

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