Question

A sample of ${}_{38}^{90}Sr$ originally had an activity of 0.500 mCi (millicurie).
What is the specific activity of the sample?

• A. $7.4\times {10}^{15}$ dis/kg-s
• B. $5.4\times {10}^{15}$ dis/kg-s
• C. $8.3\times {10}^{15}$ dis/kg-s
• D. None of these

Answer 1

The correct option is A $7.4\times {10}^{15}$ dis/kg-s

The specific activity of a radionuclide is its activity per kg of the radioactive material.

Hence, first we calculate the weight of ${}_{38}^{90}Sr$ which has an activity of 0.500 mCi
$\therefore$ $-\frac{dN}{dt}=\lambda N=\lambda \frac{w{N}_0}{A}$

$1mCi=3.7\times {10}^7$ dps

$\therefore 0.500mCi=0.5\times 3.7\times {10}^7$ dps

$\therefore 0.5\times 3.7\times {10}^7=\frac{0.693\times w\times 6.02\times {10}^{23}}{19.9\times 365\times 24\times 60\times 60\times 90}$

$\therefore w=2.5\times {10}^{-6}g=2.5\times {10}^{-9}$ kg

$\therefore$ Specific activity $=\frac{0.5\times 3.7\times {10}^7}{2.5\times {10}^{-9}}$

$=7.4\times {10}^{15}$ $dis/kg-s$