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Two radioactive elements R and S disinte...

Two radioactive elements R and S disintegrate as
`RtoP+alpha,lamda_(R)=4.5xx10^(-3)"years"^(-1)`
`StoQ+betal,lamda_(S)=3xx10^(-3)"years"^(-1)`
Starting with number of atoms of R and S in the ratio of 2:1 this ratio afte4r the lapse of three half lives of R will be

A

`3:2`

B

`1:3`

C

`1:1`

D

`2:1`

Text Solution

Verified by Experts

The correct Answer is:
C
`(lamda_(R))/(lamda_(S))=1.5`
So, the rate of disintegration of R will be 1.5 times that of S. Thus, the half life of S will be 1.5 times that of R. So, two half lives of S will be equal to the three half -lives of R
`(N_(R))/(N_(S))=(0.25)/(0.25)=1`
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