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  • The half-life of _38 ^ 90 Sr is 28 years. What is the disintegration rate of 15 mg of this isotope?

Q. 13.10 The half-life of _{38}^{90}\textrm{Sr}is 28 years. What is the disintegration rate of 15 mg of this isotope?

 

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T1/2=28 years

\\\lambda =\frac{0.693}{28\times 365\times 24\times 3600}\\ \lambda =7.85\times 10^{-10} \ decay\ s^{-1}

The number of atoms in 15 mg of  _{38}^{90}\textrm{Sr} is

N=\frac{15\times 10^{-3}\times 6.023\times 10^{23}}{90}

N=1.0038\times1020

The disintegration rate will be

\frac{\mathrm{d} N}{\mathrm{d} t}=-N\lambda

=-1.0038\times1020\times7.85\times10-10

=-7.88\times1010 s-1

The disintegration rate is therefore 7.88\times1010 decay s-1.

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