A hydrogen atom in the ground state is hit by a photon exctting electron to the thoird excerd state .The electron then drops to the second orbit. What is the frequency of radiation emitted and abserbed in the process ?

Explanation

Energy is absorbed when the electron moves from the ground state `(n= 1)` to the third excted state `(n = 4)`
First calculate the energy difference between `n = 1` and `n = 4`

Use `DeltaE_((1 rarr 4))= 2.18 xx 10^(-18) xx Z^(2) xx ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`
Here` Z = 1,n_(1) = 1, n_(2) = 4`
`:. Delta E_((1 rarr 4)) = 2.18 xx 10^(-19) xx 1^(2) xx ((1)/(1^(2)) - (1)/(4^(2))) J`
`= 2.04 xx 10^(-19)J`
This is the energy of the photon obserbed
`E_("photon") = hv = 1.04 xx 10^(-19) J` to get
`rArr v = 3.08 xx 10^(15) Hz`
Similarly , when electron jumps from `n = 4 to n= 2` the energy is emitted and is given by the same relation
Putting `n_(1)= 2 ` and `n_(2) = 4` in the expression of `Delta E` we get
`DeltaE_((4 rarr 2)) = 2.18 xx 10^(-19) xx 1^(2) xx ((1)/(2^(2)) - (1)/(4^(2))) J`
`= 4.08 xx 10^(-19) J`
The is the energy of the photon emitted
`E_("photon") = hv = 4.08 xx 10^(-19) J`
`rArr v = 6.16 xx 10^(14) Hz`