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:

$\nu=\left(\frac{1}{n^2}-\frac{1}{n^{\prime2}}\right)\frac{E_0}{h}$

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

$h=4.136\times10^{-15}eV\cdot s$

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

$\nu=\left(\frac{1}{5^2}-\frac{1}{4^2}\right)\frac{-13.6eV}{4.136\times10^{-15}eV\cdot s}=3.398\times10^{13}Hz$

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