Two radioactive substances $A$ and $B$ have decay constants $5 λ$ and $λ$ respectively. At $t = 0$ , they have the same number of nuclei. The ratio of number of nuclei of $A$ to those of $B$ will be $(1 / e^{2})$ after a time:

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Answer 1

Given : $(N_{o})_{A} = (N_{o})_{B} = N_{o}$
Using : $K t = l n \frac{N}{N _{o}}$ where, $K$ is the decay constant
$⟹$ $K t = l n N - l n N_{o}$

For substance A : $λ t = l n N_{A} - l n (N_{o})_{A}$
$∴ λ t = l n N_{A} - l n N_{o}$ .............(1)

For substance B : $5 λ t = l n N_{B} - l n (N_{o})_{B}$
$∴ 5 λ t = l n N_{B} - l n N_{o}$ .............(2)

Subtracting (2) from (1) we get:
$λ t - 5 λ t = l n \frac{N _{A}}{N _{B}}$
$- 4 λ t = l n \frac{1}{e ^{2}}$

$- 4 λ t = - 2$
$⟹ t = \frac{1}{2 λ}$

Answer 2

Given :

(N_{o})_{A} = (N_{o})_{B} = N_{o}

Using :

K t = l n \frac{N}{N _{o}}

where,

K

is the decay constant

⟹

K t = l n N - l n N_{o}

For substance A :

λ t = l n N_{A} - l n (N_{o})_{A}

∴ λ t = l n N_{A} - l n N_{o}

.............(1)

For substance B :

5 λ t = l n N_{B} - l n (N_{o})_{B}

∴ 5 λ t = l n N_{B} - l n N_{o}

.............(2)

Subtracting (2) from (1) we get:

λ t - 5 λ t = l n \frac{N _{A}}{N _{B}}

- 4 λ t = l n \frac{1}{e ^{2}}

- 4 λ t = - 2

⟹ t = \frac{1}{2 λ}

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