A hydrogen atom is in the third excited state. It make a transition to a different state and a photon is either observed or emitted. Determine the quantum number `n` of the final state and the energy energy of the photon If it is
a emitted with the shortest possible wavelength.
b emitted with the longest possible wavelength and
c absorbed with the longest possible wavelength.

Explanation

When a photon is emitted , it carries energy with it , so the final energy of the atom is less than the initial when a photon is emitted.
An atom gains energy when a photon is obsorbed. So the final quantum number is greater than the initial when a photon is obsorbed.
The largest possible energy arised when the electron jumps from the initial state `(n = 4)` to the ground state `(n_(1) = 1)` as shown by transition `A` in the figure . Therefore the quantum number of final state is `n = 1`. The energy of the nth state is given by
`E_(n) = ((Z^(2))/(n^(2)))`

The energy of the photon emitted corresponding to transition from excited state quantum number `n_(u)` to lower energy state `n_(1)` can be calculated from
`(hc)/(lambda) = `E_(u)` - E_(1) = (13.6) (1)^(2) [(1)/(n_(1)^(2)) - (1)/(n_(u)^(2))]` So ,the energy of the poton with minimum wevelength is
`Delta E = E_(4) - E_(1)`
`= (13.6) (1)^(2) ((1)/(1^(2)) - (1)/(4^(2))) = 12.75 e V`
b. The longest possible wevelength photon corresponds to minimumn energy . This happens when the electron jump from the initial state `(n_(u) = 4)` to the next lower state `(n_(1) = 3)` as shown by transition `B` in the figure. Therefore , the quantum number of the final state is `n = 3`. The energy of the photon is
`Delta E = E_(4) - E_(3) = (13.6) (1)^(2) ((1)/(3^(2)) - (1)/(4^(2))) = 0.661 e V`
c. The absorbed photon has the longest possible wavelength when its energy is the smallest , i.e., the electron jump from the initial state `(n_(1) = 4)` to the next higher state `(n_(u) = 5)` this is shown by transition `C` in the figure. Therefore , the quantum number of the final state is `n = 5`. The energy of the photon is
`Delta E = E_(5) - E_(4) = (13.6) (1)^(2) ((1)/(4^(2)) - (1)/(5^(2))) = 0.306 e V`