The ionization enthalpy of hydrogen atom is `1.312 xx 10^6 J mol^-1`. The energy required to excite the electron in the atom from `n= 1` to `n = 2` is :

Options

A. `6.56 xx 10^(5)J "atom"^(-1)`
B. `8.51 xx10^(5) J mol^(-1)`
C. `7.56 xx 10^(5)J "atom"^(-1)`
D. `9.84xx10^(5) J mol^(-1)`

Answer

D

Explanation

According to Bohr's theory
`DeltaE = Z^(2)E_(1) ((1)/(n_(f)^(2))-(1)/(n_(i)^(2)))`
When `n_(i) =1`and `n_(f) = oo`, then `DeltaE` is equal to ionization enthalpy `(IE)`.
Thus, `IE =- Z^(2)E_(1)`
or `Z^(2)E_(1) =- (IE) =- 1.312 xx 10^(6)J "atom"^(-1)`
Applying Bohr's theory, we get
`DeltaE= Z^(2)E_(1)((1)/(n_(f)^(2))-(1)/(n_(i)^(2)))`
`=- 1.312 xx 106(6)J "atom"^(-1) ((1)/(2^(2))-(1)/(1^(2)))`
`= (-1.312xx10^(6)J "atom"^(-1)) (-(3)/(4))`
`= 0.984 xx 10^(6)J "atom"^(-1)`
`= 9.840 xx 10^(5)J "atom"^(-1)`.