Question

Radioactive material 'A' has decay constant ${}^{\prime }8{\lambda }^{\prime }$ and material 'B' has decay constant ${}^{\prime }{\lambda }^{\prime }$. Initially they have same number of nuclei. After what time, the ratio of number of nuclei of material 'B' to that of 'A' will be $\frac{1}{e}$?

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

Answer 1

Answer: (B)

$\frac{1}{7\lambda }$

The following steps are to be followed for solving the query:

Step 1) Since the problem deals with decay constants, use the radioactive decay formula $N=N_0{e}^{-kt}$.

Step 2) Apply the formula for both materials A and B and find the equation $N_A$

_{and $N_B$}

Step 3) Divide $N_A$ and $N_B$ as the ratio is given. Equate it to $\frac{1}{e}$ (given). You should get $\frac{N_A}{N_B}$ = $\frac{1}{e}$ = $e^{-\left({\lambda }_A-{\lambda }_B\right)t}$

Step 4) Substitute decay constant for A and B and find the answer for t. You should get $t$ = $\frac{1}{8\lambda -\lambda }$ = $\frac{1}{7\lambda }$.