`1` mol of `He^(+)` ion is excited. Spectral analysis showed existence of `50%` ions in `3rd` level, `25%` in `2nd` level and remaining `25%` in ground state. Ionization energy of `He^(+)` is `54.4 eV`, calculate total energy evolved when all the ions retirn ground state.

Options

A. `331.13xx10^(4) J`
B. `400.14 xx10^(4) J`
C. `10^(4) J`
D. `6.66xx10^(4) J`

Answer

A

Explanation

`25%` of `He^(+)` ions are already in ground state, hence energy emitted will be from the ions present in `3rd` level and `2nd` level.
`DeltaE=(IP)_(Z)[1/n_(1)^(2)-1/n_(2)^(2)]` per ion or atom
`(DeltaE)_(3 rarr 1)=(54.4) N_(0)/2[1/1^(2)-1/3^(2)]`
for `N_(0)/2` ions falling to ground state `=54.4xx(4xxN_(0))/(9) eV`
and `(DeltaE)_(2 rarr 1)=(54.4) N_(0)/2[1/1^(2)-1/2^(2)]`
for `N_(0)/4` ions falling to ground state `=54.4xx(3xxN_(0))/(16) eV`
hence total energy `=54.4xxN_(0)[4/9+3/16]=54.4xx602xx10^(23)xx91/144 eV`
`=54.4xx6.02xx10^(23)xx91/144xx1.6xx10^(-19)J=331.13xx10^(4) J`