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Question: Two radioactive substances a and b have decay constants \(5\lambda\) and \(\lambda\) respectively At...

Two radioactive substances a and b have decay constants 5λ5\lambda and λ\lambda respectively At t = 0, they have the same number of nuclei the ration of number of nuclei of A to those of B will be (1.e)2(1.e)^{2} after a time interval

A

4λ4\lambda

B

2λ2\lambda

C

1/2λ1/2\lambda

D

1/4λ1/4\lambda

Answer

1/2λ1/2\lambda

Explanation

Solution

: Given λA=5λ,λB=λ,\lambda_{A} = 5\lambda,\lambda_{B} = \lambda,

At t=0 , (N0)A=(N0)B(N_{0})_{A} = (N_{0})_{B}

NANB=(1e)2\frac{N_{A}}{N_{B}} = \left( \frac{1}{e} \right)^{2}

According to radioactive decay, NN0=e−λt\frac{N}{N_{0}} = e^{- \lambda t}

∴NA(N0)A=e−λAt\therefore\frac{N_{A}}{(N_{0})_{A}} = e^{- \lambda}A^{t}…… (i)

NB(N0)B=e−λBt\frac{N_{B}}{(N_{0})_{B}} = e^{- \lambda}B^{t}…… (ii)

Divide (i) by (ii) we get

NANB=e−(λA−λB)torNANB=e−(5λ−λ)t\frac{N_{A}}{N_{B}} = e^{- (\lambda_{A} - \lambda_{B})t}or\frac{N_{A}}{N_{B}} = e^{- (5\lambda_{} - \lambda)t}

Or (1e)2=e−4λtor(1e)2=(1e)−4λt\left( \frac{1}{e} \right)^{2} = e^{- 4\lambda t}or\left( \frac{1}{e} \right)^{2} = \left( \frac{1}{e} \right)^{- 4\lambda t}

Or 4λt=2⇒t=24λ=12λ4\lambda t = 2 \Rightarrow t = \frac{2}{4\lambda} = \frac{1}{2\lambda}