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प्रश्न
The half-life of \[\ce{_38^90Sr}\] is 28 years. Determine the disintegration rate of its 5 mg sample.
उत्तर
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 1010s−1
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