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Standard XII

Physics

Nucleus: Common Terms

A certain rad...

Question

A certain radioactive sample is obtained by mixing equal number of nuclides of species A and species B. Half life of species A is T and that of B is 2T. After certain time, it is found that the remaining radioactive nuclei of A and B combined together are $\frac{5}{32}$ of the total number of nuclei. The number of half-lives of A that elapsed is

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Solution

The correct option is A 4
As, $N^{'} = (\frac{\frac{N}{2}}{2^{n}} + \frac{\frac{N}{2}}{2^{n / 2}})$
Where n is number of half lives elapsed for species A.
Now $\frac{N^{'}}{N} = \frac{5}{32} = [\frac{1}{2^{n + 1}} + \frac{1}{2^{\frac{n}{2} + 1}}]$
On solving, we get, n = 4

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The correct option is A 4As, N′=(N22n+N22n/2) Where n is number of half lives elapsed for species A. Now N′N=532=[12n+1+12n2+1] On solving, we get, n = 4

The correct option is A 4
As, $N^{'} = (\frac{\frac{N}{2}}{2^{n}} + \frac{\frac{N}{2}}{2^{n / 2}})$
Where n is number of half lives elapsed for species A.
Now $\frac{N^{'}}{N} = \frac{5}{32} = [\frac{1}{2^{n + 1}} + \frac{1}{2^{\frac{n}{2} + 1}}]$
On solving, we get, n = 4