Question

Radioactive material $'A'$ has decay constant $' 8 \lambda '$ and material $'B' $ has decay constant $'\lambda'$ 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 $\frac{1}{e} ?$

• A. $\frac{1}{7 \lambda} $
• B. $\frac{1}{8 \lambda} $
• C. $\frac{1}{9 \lambda} $
• D. $\frac{1}{ \lambda} $

Answer 1

The Correct Option is A

No option is correct
If we take $\frac{N_{A}}{N_{B}}=\frac{1}{e}$
Then
$\frac{N_{A}}{N_{B}}=\frac{e^{-8 \lambda t}}{e^{-\lambda t}}$
$\frac{1}{e}=e^{-7 \lambda t}$
$-1=-7 \lambda t$
$t=\frac{1}{7 \lambda}$

Answer 2

 

 

A	B
8𝜆	𝜆
N0	N0

 

 

 

 

NA/NB = 1/e

N=N0e^-𝜆t

NA = N0e^-8𝜆t……….(1)

and

NB=N0e^-𝜆t……….(2)

By solving equations 1 & 2, we get

\(\frac{N_A}{N_B}\) = \(\frac{N_0 e^- 8\lambda t}{N_0 e^-\lambda t}\)

\(\frac{N_A}{N_B}\)=\(\frac{ e^- 8\lambda t}{ e^-\lambda t}\)

\(\frac{1}{e}\) = \(\frac{1}{(e-\lambda t+e-8 \lambda t)}\)

\(\frac{1}{e}\) = \(\frac{1}{e7 \lambda t}\)

7𝜆t = 1 

t = \(\frac{1}{7\lambda}\)

So, option(A) is the correct answer.