Question

A photon is emitted from a hydrogen atom, which undergoes a transition from the n = 5 state to the n = 4 state. Find the frequency, in Hz, of the emitted photon.

Answer 1

The frequency of a photon emitted by a hydrogen atom when it passes from the state n to the state n' is:

`\\u=\left((1)/(n^2)-(1)/(n^(\prime2))\right)(E_0)/(h)`

The base energy of the Hidrogen atom is -13.6eV and the Planck's constant is equal to:

`h=4.136*10^(-15)eV\cdot s`

Replace E_0=-13.6eV, n=5 and n'=4 to find the frequency of the emitted photon:

`\\u=\left((1)/(5^2)-(1)/(4^2)\right)(-13.6eV)/(4.136*10^(-15)eV\cdot s)=3.398*10^(13)Hz`

Therefore, the frequency of the emitted photon is approximately 3.40*10^13Hz