Question

The disintegration constant of 238U is 4.87×10^(-18)s^(-1).Calculate its half-life (in years). Also calculate the number of disintegration per second from 1 gram of Uranium. It is given that Avogadro’s number = 6.02×10^23

Answer 1

Answer on Question # 84420, Physics / Atomic and Nuclear Physics

Question 1. The disintegration constant of $238U$ is $\lambda=4.87\cdot 10^{-18}s^{-1}$. Calculate its half-life $t_{1/2}$ (in years). Also calculate the number of disintegration per second from $m=1$ gram of Uranium. It is given that Avogadro’s number $N_{A}=6.02\cdot 10^{23}$.

Solution.

$t_{1/2}=\frac{\ln 2}{\lambda}=\frac{0.693}{4.87\cdot 10^{-18}}\approx 0.14\cdot 10^{18}\,s\approx 4.5\cdot 10^{9}\,y.$

The atomic mass of $238U$ is $M=238\,u$. $\nu=\frac{m}{M}=\frac{N}{N_{A}}\Rightarrow N=\frac{mN_{A}}{M}.$ The number of decays per second of a radioactive sample is

$A=\lambda N=\frac{\lambda mN_{A}}{M}\approx 12318\,s^{-1}.$

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