Energy required for excitation of electron in 1 mole H atom from ground state to `3^(rd)` excited state is 2.67 times (lesser than) dissociation energy per mole of `H_(2(g))`. Calculate the amount of energy needed to excite each H atom of `H_(2(g))` confined in 1 L at `27^(@)C` and 1 bar pressure. (R=0.083 bar litre `K^(-1)" mol"^(-1), R_(B)=1.1xx10^(-7) m^(-1)`).

Options

A. `21.87xx10^(4) J`
B. `18.21xx10^(6) J`
C. `16.32xx10^(4) J`
D. `7.789xx10^(6) J`

Answer

A

Explanation

Total mole of `H_(2) =(1xx1)/(300xx0.083)=0.040" "(because PV = nRT)`
`therefore` Total mole of H atoms `=0.040xx2=0.08`
Energy needed to excite 1 mole H atom from n =1 to n =3 is, `E=(hc)/(lamda) =hc, R_(H) [(1)/(1^(2))-(1)/(3^(2))]xxN_(A)`
`E=6.626xx10^(-34) xx3.0xx10^(8)xx1.1xx10^(7) xx(8)/(9)xx6.02xx10^(23) =11.71xx10^(5) J//mol`
Energy required for dissociation of 1 mole `H_(2)` molecules to H atoms `=11.71xx10^(5)xx2.67`
`=31.25xx10^(5)" J/mol"`
`therefore` Total energy needed = `underset("for 0.08 mole H atom")" Excitation energy" + underset("for 0.04 mole"H_(2))" Dissociation energy"`
`=11.71xx10^(5) xx0.08+31.25xx10^(5) xx0.04 =9.37xx10^(4) + 12.5xx10^(4)=21.87xx10^(4) J`