Question

The half-life of \(^{90}_{38}Sr\) is 28 years. Determine the disintegration rate of its 5 mg sample.

Answer 1

Data: T_1/2 = 28 years = 28 × 3.156 × 10⁷ s 

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

90 grams of \(^{90}_{38}Sr\) contain 6.02 × 10²³ atoms

Hence, here, N = \(\cfrac{(6.02\times10^{23})(5\times10^{-3})}{90}\)

= 3.344 × 10¹⁹ atoms 

∴ The disintegration rate = Nλ = N \(\cfrac{0.639}{T_{1/2}}\)

= \(\cfrac{(3.344\times10^{19})(0.639)}{8.837\times10^8}\)

= 2.622 × 10 disintegrations per second