Question

Radioactive material "A" has decay constant 8 lamda and
material B has decay constant lamda. Initially they have
same number of nuclei. After what time, the ratio of
number of nuclei of material B to that A will be 1/e​

Answer 1

Solution:

Radioactive material "A" has decay , $\lambda_{A} = 8 \lambda$

Radioactive material "B" has decay , $\lambda_{B} = \lambda$

After what time, the ratio of  number of nuclei of material B to that A will be 1/e​

$\frac{N_{B} }{N_{A} } = \frac{1}{e}$

$\frac{N_{0}e^{- \lambda t} }{N_{0}e^{- 8\lambda t} } = \frac{1}{e}$

$\frac{e^{- \lambda t} }{e^{- 8\lambda t} } = \frac{1}{e}$

$e^{- \lambda t} = \frac{e^{- 8\lambda t}}{e}$

$e^{- \lambda t} = e^{- 8\lambda t} * e^{-1}$

${- \lambda t} = {- 8\lambda t} {-1}$

$t = \frac{-1}{7 \lambda}$

After $\frac{1}{7 \lambda}$ time, the ratio of  number of nuclei of material B to that A will be 1/e​

Answer 2

The time taken to decay from A to B is 1/7λ

Given:

The decay constant of A = 8λ

The decay constant of B = λ

The ratio of number of nuclei of material B to that A = 1/e

Explanation:

(Number of B)/(Number of A) = 1/e

The number of radioactive is given by the formula:

N = N₀ $e^{-\lambda t}$

Now,

(N₀ $e^{-\lambda t}$)/(N₀ $e^{-8\lambda t}$) = 1/e

$e^{-\lambda t + 8\lambda t}$ = e⁻¹

On equating the powers, we get,

-λt + 8λt = -1

t (-λ + 8λ) = -1

7λt = -1

∴ t = -1/7λ