How much energy will an electron in orbit about a hydrogen atom lose in a transition from orbit # 2 to orbit # 6, where orbits are counted from the closest outward? If this lost energy is carried away by a photon, what will be its wavelength?

Thus when n2 < n1, corresponding to a transition to a 'higher' orbit of greater radius , 1/n1 > 1/n2 so the energy difference is positive--we must add energy to the system in order to accomplish the change. Conversely when n2 > n1, we are moving to a 'lower' orbit of lesser radius and energy must be removed from the system as the change occurs.

In the present case n1 < n2 so energy must be removed from the system. The mechanism by which energy is removed is the creation of a photon with the required energy.

The energy difference in the present situation is

• ( 1 / n1 - 1 / n2 ) * 13.6 eV = ( 1 / 2 - 1 / 6 ) * 13.6 eV = 4.533333 eV = 7.253334 * 10^-19 J.

The energy in electron volts is easy to relate to -- just imagine electron being accelerated across the plates of a capacitor which you have charged with the hand-held generator to the appropriate number of volts.

To calculate the wavelength of the photon we prefer to use the transition energy in Joules, since this unit is compatible with the familiar units of Plank's Constant.

Since E = h f, the frequency of the photon must be

• f = E / h = ( 7.253334 * 10^-19 J) ( / 6.62 * 10^-34 J s) = 1.09567 * 10^15 Hz.

The corresponding wavelength is `lambda = c / f = (3 * 10^8 m/s) / ( 1.09567 * 10^15 Hz ) = 273.8051 nm.

In general the transition energy is

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

The corresponding frequency is

• f = E(n1, n2) / h = ( 1 / n1 - 1 / n2 ) * [ 2 `pi^2 k^2 mE qE^4 / h^3 ]

and the wavelength is

• `lambda = c / f = c / { ( 1 / n1 - 1 / n2 ) * [ 2 `pi^2 k^2 mE qE^4 / h^2 ] } = ( n1 n2 / (n1 + n2) ) * c / [ 2 `pi^2 k^2 mE qE^4 / h^3 ].

The last expression is numerically approximated as

numerical approximation: `lambda = ( n1 n2 / (n1 + n2) ) * 91.2 nm.