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Energy of a hydrogen atom with principal...

Energy of a hydrogen atom with principal quantum number n is shown 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.

A

`1.9eV`

B

`1.5eV`

C

`0.85eV`

D

`3.4eV`

Text Solution

Verified by Experts

The correct Answer is:
A
`DeltaE=E_(3)-E_(2)=-(13.6)/(3^(2))+(13.6)/(2^(2))`
`=13.6((1)/(4)-(1)/(9))eV=1.9eV`
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