Question

A radioisotope has an initial activity of 5 millicurie. 48hours later, the observed activity is 4 millicurie. Determine: a) The half life of the radioisotope

b) The initial number of nuclides in the radioisotope.

Nb,Show workings

Answer 1

$\displaystyle A(t) = A_0 2^{-\frac{t}{T_{1/2}}}$

a) $A_0= 5, A=4, t=48 h$

$\displaystyle \frac{A}{A_0} = 2^{-\frac{t}{T}}$

$\displaystyle \frac{t}{T}= log_2{\frac{A_0}{A}}=log_2 \frac{5}{4} = 0.3219$

$\displaystyle T = \frac{48 h}{0.3219}=149.11 \, h$

b) We will need to convert everything to SI units, curie -> Bq, h -> s.

$\displaystyle A_0= N_0 \frac{ln2}{T}$

$\displaystyle N_0 = \frac{A_0 T}{ln 2}=\frac{5 \cdot 10^{-3}\cdot 3.7⋅10^{10} \cdot 149.11 \cdot 3600}{0.6931} =14. 328 \cdot 10^{13}= 1.4328 \cdot 10^{14}$