Two radioactive nuclei A and B have their disintegration constant λA and λB respectively initially NA and NB number of nuclei are taken then the time after which their undisintegrated nucleus are same is:

Question

Two radioactive nuclei A and B have their disintegration constant λA and λB respectively initially NA and NB number of nuclei are taken then the time after which their undisintegrated nucleus are same is: (A) (1/λ A +λ B ) ln ( N B / N A ) (B) (λA λB/λA-λB) ln (NB/NA) (C) (1/λ A -λ B ) ln ( N A / N B ) (D) (1/λA-λB) ln (NB/NA)

Tardigrade Admin · Accepted Answer

N A e -λA t= N B e -λ B t e (λ B -λA) t=( N B / N A ) (λB-λA) t= ln (NB/NA) t =(1/λ B -λ A ) ln ( N B / N A )

Q. Two radioactive nuclei $A$ and $B$ have their disintegration constant $λ_{A}$ and $λ_{B}$ respectively initially $N_{A}$ and $N_{B}$ number of nuclei are taken then the time after which their undisintegrated nucleus are same is:

$\frac{1}{λ _{A} + λ _{B}} ln \frac{N _{B}}{N _{A}}$

$\frac{λ _{A} λ _{B}}{λ _{A} - λ _{B}} ln \frac{N _{B}}{N _{A}}$

$\frac{1}{λ _{A} - λ _{B}} ln \frac{N _{A}}{N _{B}}$

$\frac{1}{λ _{A} - λ _{B}} ln \frac{N _{B}}{N _{A}}$

$N_{A} e^{- λ_{A} t} = N_{B} e^{- λ_{B} t}$
$e^{(λ_{B} - λ_{A}) t} = \frac{N _{B}}{N _{A}}$
$(λ_{B} - λ_{A}) t = ln \frac{N _{B}}{N _{A}}$
$t = \frac{1}{λ _{B} - λ _{A}} ln \frac{N _{B}}{N _{A}}$

Q. Two radioactive nuclei A and B have their disintegration constant λA​ and λB​ respectively initially NA​ and NB​ number of nuclei are taken then the time after which their undisintegrated nucleus are same is:

λA​+λB​1​lnNA​NB​​

λA​−λB​λA​λB​​lnNA​NB​​

λA​−λB​1​lnNB​NA​​

λA​−λB​1​lnNA​NB​​