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 $3 \, In \left(\frac{A_{0}}{6000}\right) = 2 \,In \left(\frac{A_{0}}{3000}\right)$
$\therefore \:\:\: \left(\frac{A_{0}}{6000}\right)^3 = \left(\frac{A_{0}}{3000}\right)^2$
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 $A_0 = 24000$