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There are two radioactive substances A and B. Decay constant of B is two times that of A. Initially, both have equal number of nuclei. After n half lives of A, rate of disintegration of both are equal. The value of n is

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a

4

b

2

c

1

d

5

answer is C.

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Detailed Solution

$\begin{array}{l} λ_{B} = 2 λ_{A} {nT}_{A} = \frac{N_{0}}{2^{n}} = N_{A} \\ T_{B} = \frac{1}{2} T_{A} \frac{N_{0}}{2^{2 n}} = N_{B} λ_{A} N_{A} = λ_{B} N_{B} λ_{A} \frac{N_{0}}{2^{n}} = 2 λ_{A} \frac{N_{0}}{2^{2 n}} \\ Give \frac{2^{2 n}}{2^{n}} = 2 2^{2 n - n} = 2^{1} n = 1 \end{array}$

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