A hydrogen atom is in an excited state c...

A hydrogen atom is in an excited state corresponding to an energy level of -0.85 eV. The frequency of the photon emitted if it comes down to the ground state (-13.6 eV) in a single jump is

`3.077xx10^(15)` Hz

`30.77xx10^(15)` Hz

`307.7xx10^(15)` Hz

`3077xx10^(15)` Hz

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The correct Answer is:

Energy of photon `=hv=E_(n)-E_(p)`
`=-0.85+13.6=12.75` eV
`:. v=(12.75xx1.6xx10^(-19))/(6.63xx10^(-34))`
`=3.077xx10^(15)` Hz

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