Disintegration rate

Asked by  | 18th 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.

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Answered by  | 18th Feb, 2010, 11:56: AM

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