Polonium (`_84^210`Po`) 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`

Answer

A, B, D

Explanation

`_84^210Porarr_82^206Pb+_2^4He`
`Deltam=0.00564` amu
Energy liberated per reaction `=(Deltam)931MeV=8.4xx10^-13J`
Electrical energy produced `=8.4xx10^-14J`
Let m g of `^210Po` is required to produce the desired energy.
`N=(m)/(210)xx6xx10^23`
`lambda=(0.693)/(t_(1//2))=0.005` per day
`(-(dN)/(dt))=lambdaN=((0.005)(6xx10^23m))/(210)`per day
`:.` Electrical energy produced per day
`=((0.005)(6xx10^23m))/(210)xx8.4xx10^-14J`
This is equal to `1.2xx10^7J` (given)
`:.` `m=10g`
Activity at the end of 693 days is
`R=(0.005xx6xx10^23xx10)/(210)=(10^21)/(7)` per day = `R_0(1/2)^n`
Here, n=number of half-lives=(693)/(138.6)=5
`:.` `R_0=R(2)^5=32xx10^21/7=4.57xx10^21` per day