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Question:
Some nuclei of a radioactive material are undergoing radioactive decay. The time gap between the instances when a quarter of the nuclei have decayed and when half of the nuclei have decayed is given as :
(where $\lambda$ is the decay constant)
Correct Option: , 4
Solution:
$\frac{3 \mathrm{~N}_{0}}{4}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{1}}$
$\frac{\mathrm{N}_{0}}{2}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{2}}$
$\ln (3 / 4)=-\lambda t_{1} \ldots . .$ (i)
$\ln (1 / 2)=-\lambda t_{2} \quad \ldots .$ (i)
$\ln (3 / 4)-\ln (1 / 2)=\lambda\left(t_{2}-t_{1}\right) \quad \ldots .$ (i)
$\Delta \mathrm{t}=\frac{\ln (3 / 2)}{\lambda}$
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