Question

A and B are two radioactive elements having initial activity of$10\mathrm{curie}\mathrm{and}20\mathrm{curie}$.If A has twice the number of moles as compared to B.Calculate the decay constant ${\mathrm{\lambda }}_{\mathrm{A}}\mathrm{and}{\mathrm{\lambda }}_{\mathrm{B}}$.

• A. $\left(10,5\right)$
• B. $\left(5,20\right)$
• C. $\left(20,10\right)$
• D. $\left(50,100\right)$

Answer 1

The correct option is B

$\left(5,20\right)$

Explanation for the correct options:

B. $\left(5,20\right)$

Step 1:

• The activity of an element is defined as a number of nuclei that get decayed with time.
• Activity is denoted by ‘A’.
• $\mathrm{A}=\frac{-\mathrm{dN}}{\mathrm{dT}}$

Step 2:

• Given,A_A=$10\mathrm{curies}$
• A_B=$20\mathrm{curies}$
• n_A=2n_B; let n_B=x then n_A=$2\mathrm{x}$
• $\frac{\mathrm{dN}}{\mathrm{dt}}=-\mathrm{\lambda N}$$=\left(\frac{-\mathrm{dN}}{\mathrm{dt}}\right)\times \frac{1}{\mathrm{N}}$

Step 3:

• For A=${\lambda }_A=10\times \frac{1}{2x}$$=5x$
• For B=${\lambda }_B=20\times \frac{1}{x}=\frac{20}{x}$

Step4:

• ${\lambda }_A:{\lambda }_B=5:20$

Hence,option(B)is correct,the decay constant ${\mathrm{\lambda }}_{\mathrm{A}}\mathrm{and}{\mathrm{\lambda }}_{\mathrm{B}}$ is $\left(5,20\right)$.