The half life of \( {}^{90}_{38}\mathrm{Sr} \) is 28 years. The disintegrating rate of 15 mg of this isotope is (Given 1 year \( = 3.15\times10^{7}\,\mathrm{s} \)):
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 7.85\times10^{10}\,\mathrm{Bq} \)
The disintegration rate (activity) is \( R = \lambda N \), where the decay constant is \( \lambda = 0.693/T_{1/2} \) and \( N \) is the number of nuclei present.
\( \displaystyle T_{1/2} = 28 \times 3.15\times10^{7}\,\mathrm{s} = 8.82\times10^{8}\,\mathrm{s} \)
\( \displaystyle \lambda = \frac{0.693}{8.82\times10^{8}\,\mathrm{s}} = 7.86\times10^{-10}\,\mathrm{s^{-1}} \)
One mole (90 g) of strontium-90 contains \( 6.02\times10^{23} \) nuclei, and \( 15\,\mathrm{mg} = 15\times10^{-3}\,\mathrm{g} \).
\( \displaystyle N = \frac{15\times10^{-3}}{90} \times 6.02\times10^{23} = 1.0\times10^{20} \)
\( \displaystyle R = \lambda N = 7.86\times10^{-10} \times 1.0\times10^{20} \approx 7.86\times10^{10}\,\mathrm{Bq} \)
This is \( 7.85\times10^{10}\,\mathrm{Bq} \) to within the rounding of the constants.
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Exercises – …
