A photon of 77.76066 nm electromagnetic radiation encounters an electron in the n = 2 orbital of a hydrogen atom, and causes it to 'jump' to the n = 3 orbital. What will be the wavelength of a photon which carries away any excess energy from the collision?

As derived in a preceding problem we see that the difference in orbital energies from orbital n1 to orbital n2 is

• ( 1 / n1 - 1 / n2 ) * 13.6 eV.

The energy difference is

• ( 1 / n1 - 1 / n2 ) * 13.6 eV = ( 1 / 2 - 1 / 3 ) * 13.6 eV = 2.266667 eV = 3.626667 * 10^-19 J.

Since n2 > n1 this energy is positive. The electron must gain energy from the photon.

The photon initially has energy E = hf = h ( c / `lambda):

• initial photon energy = h c / `lambda = 6.62 * 10^-34 J s * (3 * 10^8) m/s / ( 77.76066 * 10^-9 m) = 25.53991 * 10^-19 J.

The photon loses the 3.626667 * 10^-19 J of energy, ending up with

• phinal foton energy = ( 25.53991 - 3.626667) * 10^-19 J = 21.91324 * 10^-19 J.

The final photon wavelength is therefore

• final photon wavelength = c / f = c / ( E / h ) = h c / E = 6.62 * 10^-34 J s * (3 * 10^8 m/s) / ( 21.91324 * 10 ^ -19 J ) = 234915.3 nm.

In general the transition energy is

• E(n1, n2) = ( 1 / n1 - 1 / n2 ) * [ 2 `pi^2 k^2 mE qE^4 / h^2 ].

A photon with wavelength `lambda will have energy

• photon energy E = hf = h c / `lambda

so that after giving up energy E(n1, n2) the photon will carry off the remaining energy

• final photon energy Efinal = h c / `lambda - E( n1, n2 ) = h c / `lambda - ( 1 / n1 - 1 / n2 ) * [ 2 `pi^2 k^2 mE qE^4 / h^2 ].

This photon will have frequency and wavelength corresponding to this energy; the actual algebraic expression isn't particularly enlightening and will not be included here.