Question

A 280 day old radioactive substance shows an activity of 6000 dps, 140 days later its activity becomes 3000 dps. What was its initial activity?

• A. 20000 dps
• B. 24000 dps
• C. 12000 dps
• D. 6000 dps

Answer 1

Answer: (B)

For radioactive disintegration,
$\lambda =\frac{1}{t}In\frac{N_0}{N}=\frac{1}{t}In\left(\frac{A_0}{A}\right)$
or $\lambda =\frac{1}{280}In\left(\frac{A_0}{6000}\right)$
Also $\lambda =\frac{1}{(280+140)}In\left(\frac{A_0}{3000}\right)$
$\therefore \frac{1}{280}In\left(\frac{A_0}{6000}\right)=\frac{1}{420}In\left(\frac{A_0}{3000}\right)$
or $3In\left(\frac{A_0}{6000}\right)=2In\left(\frac{A_0}{3000}\right)$
$\therefore {\left(\frac{A_0}{6000}\right)}^3={\left(\frac{A_0}{3000}\right)}^2$
or $\frac{A_0^3}{A_0^2}=\frac{(6000{)}^3}{(3000{)}^2}$ or $A_0=\frac{6\times 6\times 6\times 10^9}{3\times 3\times 10^6}$
or $A_0=24\times 10^3$ or $A_0=24000$