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A sample of radioactive substance decays simultaneously by two processes A and B with half lives 1/2h and 1/4h respectively. For the first half hour it decays with process A, next one hour with process B, and for further half an hour with both A and B. If originally, there are N₀ nuclei, then the number nuclei after 2h of such decay

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$\frac{N_{0}}{2^{8}}$

$\frac{N_{0}}{2^{4}}$

$\frac{N_{0}}{2^{6}}$

$\frac{N_{0}}{2^{5}}$

answer is A.

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Detailed Solution

$\frac{N}{N_{0}} = \frac{1}{2^{n}} = \frac{1}{2^{t / T}}$

and in parller decay average half life

$T = \frac{T_{1} T_{2}}{T_{1} + T_{2}}$

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