Disintegration rate

(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 /T_1/2)(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.