CBSE Class 12-science Answered
Disintegration rate
Asked by | 18 Feb, 2010, 10:52: AM
Expert Answer
(b) 2.13 Ci
Number of particles of 90Sr38 in a 15 mg sample = 15x10-3x6.023x1023/90
28 years = 28x365x24x3600 sec
Disintegration rate = -dN/dt = λN
= (ln 2 /T1/2)(N)
= (ln2/(28x365x24x3600))(15x10-3x6.023x1023/90)
= 7.8782 x 1010 disintegration/sec
= 7.8782 x 1010/(3.7 x 1010) ...............3.7 x 1010 disintegration/sec = 1 Ci
= 2.13 Ci.
Regards,
Team,
TopperLearning.
Answered by | 18 Feb, 2010, 11:56: AM
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