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energy E of a hydrogen atom with princip...

energy E of a hydrogen atom with principal 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 `n=2` state of hydrogen is approximately

A

`1.9 eV`

B

`2.3 eV`

C

`3.4 eV`

D

`4.5 eV`

Text Solution

Verified by Experts

The correct Answer is:
A
`13.6 ((1)/(2^(2)) - (1)/(3^(2))) eV = 1.9 eV`
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