`1.8 g` of hydrogen is excite by irradiation. The study of spectra indicated that `27 %` of the atoms are in the first excite state, `15 %` of the aton in the second excited, and the rest in the ground state. The ground stat, ionization energy of hydrogen atom is `21.4 xx 10^(-12)`ergs.
The total amount of energy that would be evolved when all the atoms neutron to the ground state is

Options

A. `782 k J`
B. `978 k J`
C. `19.63 xx 10^(11) erg`
D. `97.87 xx 10^(11) erg`

Answer

A

Explanation

Amount of energy that would be evolved `= U_(i) - U_(f)`
`U_(i) = N_(1) E_(1) + N_(2) E_(2)`
`= - 1.61 xx 10^(23) xx (21 xx 10^(-12))/(3^(2))
- (2.9 xx 10^(23) xx 21.7 xx 10^(-12))/2^(2)`
`= - 3.88 xx 10^(11) - 15.75 xx 10^(11)`
`= - 19.63 xx 10^(11)` erg
`U_(f) = (N_(1) + N_(2)) E_(0)`
`= - (1.61 xx 10^(23) + 2.9 xx 10^(23)) 21.7 xx 10^(-12)`
`= - 97.87 xx 10^(11)`erg
Energy evolved
`U_(i) - U_(f) = (- 19.63 xx 10^(11) + 97.87 xx 10^(11))`
`= 78.237 xx 10^(11) = 782 k J`