Polonium `(._84^210Po)` emits `._2^4He` particles and is converted into lead (`._82^206Pb`). This reaction is used for producing electric power in a space mission. `Po^210` has half-life of `138.6` days. Assuming an efficiency of `10%` for the thermoelectric machine, how much `.^210Po` is required to produce `1.2xx10^7J` of electric energy per day at the end of `693 days`. Also find the initial activity of the material.
Given : Masses of nuclei
`.^210Po=209.98264` amu, `.^206Pb=205.97440` amu, `._2^4He=4.00260` amu,
`1 am u=931 MeV//c^2` and Avogadro's number` = 6xx10^23//mol`

Explanation

`._(84)^(210)Po rarr._(82)^(206)Pb +_(2)^(4)He`
Mass defect: `Deltam = 209.98264-(205.97440+4.00260)`
`= 0.00564 amu`
Equivalent energy `E = 0.00564 xx 931MeV = 5.25MeV`
`= 5.25xx1.6 xx 10^(-13) = 8.4xx10^(-13) J`
As efficiency is `10%`, equivalent energy `= 0.1xx8.4xx10^(-13)`
`= 0.84xx10^(-13) J`
Total amount of energy required `=1.2 xx 10^(7)J`
Number of reactions required per day
`= (1.2xx10^(7))/(0.84xx10^(-13)) = (1)/(7)xx10^(21)`
Activity `A= lambdaN`
`A = (ln2)/(t_(1//2))N`
`(1)/(7)xx10^(21) = (0.693)/(138.6)N`
`N = (200)/(7)xx10^(21)`
`N = (m)/(At.wt.)N_(A)`
`(200)/(7) xx10^(21) = (m)/(210)xx6xx10^(23)`
`m=1g`
`t = 693` days, `t_(1//2) = 138.6` days
Number of half-lives `n = (t)/(t_(1//2)) = (693)/(138.6) = 5`
Initial mass of `P_(0)` required `(m)/(m_(0)) = (N)/(N_(0)) = ((1)/(2))^(n) = ((1)/(2))^(5) = (1)/(32)`
`m_(0)=32 m = 32 g`
Initial number of atoms, `N_(0) = 32N = (6400)/(7)xx10^(21)`
Initial activity
`A_(0) = lambdaN_(0) = (0.613)/(138.6)xx(6400)/(7)xx10^(21) = 4.57xx10^(21)//day`