A hydrogen atom initially in the ground level absorbs a photon, Which excites it to n=4 level. Determine the wavelength and frequency of photon.

Explanation

For ground level `n_(1)=1`
Let `E_(1)` be the energy of this leve it is known that `E_(1)` is related with `n_(1)` as:
`E_(1)=(-13.6)/n_(1)^(2) eV`
`=(-13.6)/I^(2)=-13.6 eV`
The atom is excited to a higher level `n_(2)=4`
Let `E_(2)` be the energy of this level
`:.E_(2)=(-13.6)/n_(2)^(2) eV`
`=(-13.6)/4^(2)=-(13.6)/16 eV`
The amount of energy absorbed by the photon is given as:
`E=E_(2)-E_(1)`
`(-13.6)/16-(-(13.6)/1)`
`=(13.6xx15)/16 eV`
`=(13.6xx15)/16 xx1.6xx10^(-19)=2.04xx10^(-18)J`
For a photon of wavelength `lambda` the expression of energy is written as:
`E=(hc)/lambda`
Where,
h=Planck's constant `=6.6xx10^(-34)Js`
c=Speed of light `=3xx10^(8)m//s`
`:. lambda=(hc)/E`
`=(6.6xx10^(-34)xx3xx10^(8))/(2.04xx10^(-18))`
`=9.7xx10^(-8)m=97nm`
And, frequency of a photon is given by the relation
`v=c/lambda`
`=(3xx10^(8))/(9.7xx10^(-8)) approx 3.1xx10^(15)Hz`
Hence, the wavelength of the photon is 97 nm while the frequency is `3.1xx10^(15)Hz`