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The half-life of X9038X290238SrX3890X238290Sr\ce{_38^90Sr} is 28 years. Determine the disintegration rate of its 5 mg sample. - Physics

Numerical

The half-life of \[\ce{_38^90Sr}\] is 28 years. Determine the disintegration rate of its 5 mg sample.

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Solution

Data: T1/2 = 28 years = 28 x 3.156 x 107 s

= 8.837 x 108 s, M = 5mg = 5 x 10-3 g

90 grams of \[\ce{_38^90Sr}\] contain 6.02 x 1023 atoms

Hence, here, N = `((6.02 xx 10^23)(5 xx 10^-3))/90`

= 3.344 × 1019 atoms

∴ The disintegration rate = `"N"lambda = "N"0.693/("T"_(1//2))`

`= ((3.344 xx 10^19)(0.693))/(8.837 xx 10^8)`

= 2.622 x 1010 disintegrations per second

Concept: Atomic Nucleus
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APPEARS IN

Balbharati Physics 12th Standard HSC Maharashtra State Board
Chapter 15 Structure of Atoms and Nuclei
Exercises | Q 17 | Page 343
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