Question

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

• A. $20000dps$
• B. $24000dps$
• C. $120000dps$
• D. $6000dps$

Answer 1

Answer: (B)

$24000dps$

$I\left(t\right)=I_o{e}^{-\lambda t}$
$t=280$ days
$6000=I_o{e}^{-280\lambda }$ -------------(1)
$3000=I_o{e}^{-420\lambda }$ -------------(2)

Solving (1) and (2):
Two equations, two variables
$I_o=24000dps$