Question 42.31: Consider an initially pure 3.4 g sample of ^67Ga, an isotope......

(a) The decay rate is given by R = λN, where λ is the disintegration constant and N is the number of undecayed nuclei. Initially, R=R_0=\lambda N_0 , where N_0 is the number of undecayed nuclei at that time. One must find values for both N_0 and λ . The disintegration constant is related to the half-life  T_{1/ 2} by

\lambda=(\ln 2) / T_{1 / 2}=(\ln 2) /(78 h )=8.89 \times 10^{-3} h ^{-1} .

If M is the mass of the sample and m is the mass of a single atom of gallium, then N_0 = M/m. Now,

m=(67 \,u )\left(1.661 \times 10^{-24} \,g / u \right)=1.113 \times 10^{-22} \,g

and

N_0=(3.4 \,g ) /\left(1.113 \times 10^{-22} g \right)=3.05 \times 10^{22} .

Thus,

R_0=\left(8.89 \times 10^{-3}\, h ^{-1}\right)\left(3.05 \times 10^{22}\right)=2.71 \times 10^{20} \,h ^{-1}=7.53 \times 10^{16} s ^{-1}

(b) The decay rate at any time t is given by

R=R_0 e^{-\lambda t}

where R_0 is the decay rate at t = 0. At t = 48 h, \lambda t=\left(8.89 \times 10^{-3}\, h ^{-1}\right)(48\, h )=0.427 and

R= c 7.53 \times 10^{16} s ^{-1} \,he ^{-0.427}=4.91 \times 10^{16} s ^{-1} \text {. }