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Uranium .(92)^(238)U is an ustable nucle...

Uranium `._(92)^(238)U` is an ustable nucleus. It decays to Thorium `._(92)^(238)Th`, which is again an unstable nucleus which further decays to `._(91)^(234)Pa`. Let `._(92)^(238)U` be called `A` of decay constant `lambda_(1)` and `._(90)^(234)Th` is called as `B` of decay constant `lambda_(2)` and stable nuclei `._(91)^(234)Pa` be called as `C`. Here `A` is called parent nucleus and `B` is called daughter nucleus of `A`. Any two adjacent nuclei may be consider parent or daughter nuclei `A, B` and `C` respectively at time `'t'`.
Then we can write `Aoverset(lambda_(1))rarrBoverset(lambda_(2))rarrC`
Rate of disintergration of `A=(dN_(1))/(dt)=-lambda_(1)N_(1)`
Rate of disintergration of `B=(dN_(2))/(dt)=lambda_(1)N_(1)-lambda_(2)N_(2)`
Rate of formation of nuclei C is equal to
`(dN_(3))/(dt)=-lambda_(2)N_(2)`
If at `t=0`, there are `N_(0)` number of nuclei of `A` where as nuclei `B` and `C` are absent in the sample Answer the following questions ltbgt Numeber of the nuclei of nuclei C at time `t` is

A

`(N_0lambda_1)/(lambda_2-lambda_1)(e^(lambda_1t)-e^(lambda_2t))`

B

`(N_0lambda_2)/(lambda_2-lambda_1)(e^(-lambda_1t)-e^(-lambda_2t))`

C

`(N_0lambda_1)/(lambda_2-lambda_1)(e^(lambda_1t)-e^(lambda_2t))`

D

`(N_0lambda_2)/(lambda_2-lambda_1)(e^(lambda_1t)+e^(lambda_2t))`

Text Solution

Verified by Experts

The correct Answer is:
C
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Uranium ._(92)^(238)U is an ustable nucleus. It decays to Thorium ._(92)^(238)Th , which is again an unstable nucleus which further decays to ._(91)^(234)Pa . Let ._(92)^(238)U be called A of decay constant lambda_(1) and ._(90)^(234)Th is called as B of decay constant lambda_(2) and stable nuclei ._(91)^(234)Pa be called as C . Here A is called parent nucleus and B is called daughter nucleus of A . Any two adjacent nuclei may be consider parent or daughter nuclei A, B and C respectively at time 't' . Then we can write Aoverset(lambda_(1))rarrBoverset(lambda_(2))rarrC Rate of disintergration of A=(dN_(1))/(dt)=lambda_(1)N_(1) Rate of disintergration of B=(dN_(2))/(dt)=lambda_(1)N_(1)-lambda_(2)N_(2) Rate of formation of nuclei C is equal to (dN_(3))/(dt)=lambda_(2)N_(2) If at t=0 , there are N_(0) number of nuclei of A where as nuclei B and C are absent in the sample Answer the following questions Number of nuclei of B at any time t is

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