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Two radioactive nuclei P and Q, in a giv...

Two radioactive nuclei `P` and `Q`, in a given sample decay into a stable nucleus `R`. At time `t = 0`, number of `P` species are `4 N_0` and that of `Q` are `N_0`. Half-life of `P` (for conversation to `R`) is `1mm` whereas that of `Q` is `2 min`. Initially there are no nuclei of `R` present in the sample. When number of nuclei of `P` and `Q` are equal, the number of nuclei of `R` present in the sample would be :

A

`3 N_0`

B

`(9 N_0)/(2)`

C

`(5 N_0)/(2)`

D

`2 N_0`

Text Solution

Verified by Experts

The correct Answer is:
B

(b) Initially `p rarr 4 N_0`
`Q rarr N_0`
Half-life `T_p rarr 1 min`
`T_Q rarr 2 min`
Let after time `t` number of nuclei of `P` and `Q` are equal,
i.e., `(4 N_0)/(2^(t//1)) = (N_0)/(2^(t//2))`
`4 = 2^(t//2)`
`2^2 = 2^(t//2) rArr (t)/(2) = 2`
`t = 4 min`
Deactivate nucleus or nuclei of `R`
=`(4 N_0 - (4 N_0)/(2^4)) + (N_0 - (N_0)/(2^2))`
=`4 N_0 (N_0)/(4) + N_0 - (N_0)/(4) = 5 N_0 - (N_0)/(2) = (9)/(2) N_0`.
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Knowledge Check

  • Two radioactive nuclei, P and Q, are present in a given sample and both of them decay into a stable nuclei R with different decay constants. At time t = 0, the number of active nuclei of P are 4N_(0) and that of Q are N_(0) . Half-life of P is 1 minute, where as that of Q is 2 minutes. Initially, there are no nuclei of R present in the sample. When the number of nuclei of P and Q become equal, then the number of nuclei of R present in the sample would be

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    D
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  • We have two radioactive nuclei A and B . Both convert into a stable nucleus C . Nucleus A converts into C after emitting two two alpha -particles and three beta -particles. Nucleus B converts into C after emitting one. alpha -particle anf five beta -particles. A time t = 0 , nuclei of A are 4N_(0) and that of B are N_(0) . Half-life of A (into the conversion of C ) is 1min and that of B is 2min . Initially number of nuclei of C are zero. What are number of nuclei of C , when number of nuclei of A and B are equal ?

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  • If N_(0) is the number of radioactive nuclei initially present , then the number of nuclei remaining undecayed at the end of nth half life is

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    `2^(-n) N_(0)`
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