1.

The half life of a radioactive sample 38Sr90 is 2 years. Calculate the rate of disintegration of 15 mg of this isotope. Given Avogadro number = 6.023 x 1023.

Answer»

T1/2 = 28 years = 28 x 365 x 24 x 600 = 8.83 x 108 seconds 

90g \(Sr_{38}^{90}\) contains 6.023 x 1023 atoms 

λ = \(\frac{0.693}{T_{1/2}}\)

Decay constant λ = 7.85 x 1010 x 1.004 x 1.004 × 1010 

Rate of disintegration R= λN 

R = 7.88 x 1010 Bq



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