Radioactive material 'A' has decay constant ' 8 λ ' and material 'B' has decay constant 'λ' Initially they have same number of nuclei. After what time, the ration of number of nuclei of material of material 'B' to that 'A' will be e ?

Question

Radioactive material 'A' has decay constant ' 8 λ ' and material 'B' has decay constant 'λ' Initially they have same number of nuclei. After what time, the ration of number of nuclei of material of material 'B' to that 'A' will be e ? (A) (1/7 λ)  (B) (1/8 λ)  (C) (1/9 λ)  (D) (1/ λ)

Tardigrade Admin · Accepted Answer

N A = N A 0 e (-8 λ) t N B = N B 0 e -λ t N A 0= N B 0 ⇒ ( N B / N A )= e ⇒ ( e -λ t / e -8 λ t )= e 1 ⇒ e 7 λ t = e 1 7 λ t =1 t =(1/7 λ)

Q. Radioactive material $^{'} A^{'}$ has decay constant $^{'} 8 λ^{'}$ and material $^{'} B^{'}$ has decay constant $^{'} λ^{'}$ Initially they have same number of nuclei. After what time, the ration of number of nuclei of material of material $^{'} B^{'}$ to that $^{'} A^{'}$ will be $e ?$

$\frac{1}{7 λ}$

$\frac{1}{8 λ}$

$\frac{1}{9 λ}$

$\frac{1}{λ}$

$N_{A} = N_{A_{0}} e^{(- 8 λ) t}$
$N_{B} = N_{B_{0}} e^{- λ t}$
$N_{A_{0}} = N_{B_{0}}$
$\Rightarrow \frac{N _{B}}{N _{A}} = e$
$\Rightarrow \frac{e ^{- λ t}}{e ^{- 8 λ t}} = e^{1}$
$\Rightarrow e^{7 λ t} = e^{1}$
$7 λ t = 1$
$t = \frac{1}{7 λ}$

Q. Radioactive material ′A′ has decay constant ′8λ′ and material ′B′ has decay constant ′λ′ Initially they have same number of nuclei. After what time, the ration of number of nuclei of material of material ′B′ to that ′A′ will be e?

7λ1​

8λ1​

9λ1​

λ1​