Energy `E` of a hydrogen atom with principle quantum number `n` is given by `E = (-13.6)/(n^(2)) eV`. The energy of a photon ejected when the electron jumps from `n = 3` state to `n = 2` state of hydrogen is approximately

Options

A. `1.5 eV`
B. `0.85 eV`
C. `3.4 eV`
D. `1.9 eV`

Answer

D

Explanation

Given `E_(n) =-(13.6 eV)/(n^(2))`
Energy of photon ejected when electron jumps from `n = 3` state to `n = 2` state is given by
`DeltaE = E_(3) - E_(2)`
`:. E_(3) =-(13.6)/((3)^(2))eV =-(13.6)/(4)eV` `E_(2) =- (13.6)/((2)^(2))eV =- (13.6)/(4)eV`
SO, `DeltaE = E_(2) - E_(2) =- (13.6)/(9) - (-(13.6)/(4))`
`= 1.9eV` (approximately)
`E_(3 rarr 2) =- 3.4 -(-1.51) =- 1.89eV`
`rArr |E_(3 rarr 2)| ~~ 1.9 eV`