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NCERT Class XII Chapter
Nuclei
Questions With Solutions

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Question : 10
Total: 31
The half-life of
90
‌
38
Sr is 28 years. What is the disintegration rate of 15 mg of this isotope?
Solution:  
Given, T1/2 = 28 years = 28 × 3.154 × 107 s
Mass, m = 15 mg = 0.015 g
Number of atoms in 0.015 g sample of
90
‌
38
Sr ,
N =
m
M
 × ×Avogadro's number =
0.015×.023×1023‌atoms
90
Activity of the sample,
R = λN =
0.693
T1∕2
 N =
0.693×0.015×6.023×1023
28×3.154×107×90
= 7.877 × 1010 disintegration/sec = 7.877 × 1010 Bq
=
7.877×1010
3.7×1010
 Ci = 2.13 Ci
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