Question

12. A radioactive sample takes 3.5 minutes to reduce its activity from 2.0 10 Bq to 7.5-10 Bq. Calculate the half life of the sample. [2 marks]

Answer 1

Half life of the sample = T_1/2

Decay constant = $\lambda$

$\lambda = \frac{ln(2)}{T_{1/2}}$

Initial activity of the radioactive sample = N₀ = 2.0 x 10⁵ Bq

Activity of the radioactive sample after 3.5 minutes = N = 7.5 x 10³ Bq

Time period = T = 3.5 min

$N = N_{0}e^{-\lambda T}$

$7.5\ast 10^{3} = (2.0\ast 10^{5})e^{-\lambda T}$

0,0375 = e-7

Taking natural log on both sides,

$ln(0.0375) = -\lambda T$

$ln(0.0375) = -(\frac{ln(2)}{T_{1/2}})(3.5)$

T_1/2 = 0.739 min

Half life of the sample = 0.739 min