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) 12000dps (D) 6000 dps

Answer 1

Answer on Question #76810, Physics / Atomic and Nuclear Physics

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) 12000dps (D) 6000 dps.

Solution:

$$\lambda = \frac{1}{t} \ln \left(\frac{A_0}{A}\right)$$ $$\lambda = \frac{1}{280} \ln \left(\frac{A_0}{6000}\right)$$

So,

$$\lambda = \frac{1}{280 + 140} \ln \left(\frac{A_0}{3000}\right)$$ $$3 \ln \left(\frac{A_0}{6000}\right) = 2 \ln \left(\frac{A_0}{3000}\right)$$ $$\left(\frac{A_0}{6000}\right)^3 = \left(\frac{A_0}{6000}\right)^2$$

We get

$$\frac{A_0^3}{A_0^2} = \frac{6000^3}{3000^2}$$ $$A_0 = 24 \times 10^3 = 24000$$

Answer: 24000

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