Question

52. Initially:
[
Nleft({ }_{1} mathrm{H}^{3}+{ }_{1} mathrm{H}^{1}right)=frac{10}{8} times 2 times 6 times 10^{23}=frac{20}{3} times 10^{23}
]
( Rightarrow quad 1+frac{Nleft({ }_{1} mathrm{H}^{1}right)}{Nleft({ }_{1} mathrm{H}^{3}right)}=frac{20 times 10^{23}}{3 Nleft({ }_{1} mathrm{H}^{3}right)} )
( Rightarrow quad 1+frac{1}{8 times 10^{-18}}=frac{20 times 10^{23}}{3 Nleft(1 mathrm{H}^{3}right)} approx 1.25 times 10^{17} )
( Rightarrow quad Nleft({ }_{1} H^{3}right)=frac{20 times 10^{23}}{3 times 1.25 times 10^{17}}=5.33 times 10^{6} )
( Rightarrow quad k t=ln frac{N_{0}}{N} )
( Rightarrow quad frac{ln 2}{12.3} times 40=ln frac{5.33 times 10^{6}}{N} )
( Rightarrow quad N=5.6 times 10^{5} )

# 52. The nucleidic ratio, H to H in a sample of water is 8.0 x 10-18: 1. Tritium undergoes decay with a half-life period of 12.3 yr. How many tritium ators would 10.0 g of such a sample contain 40 yr after the original sample is collected 1992, 4MD

Solution