Question

Calculate the fequency in Hz of electromagnetic radiation released when an electron makes the transition from n=5 to n=2 ?

Answer 1

The frequency is `\nu=691.11` THz.

Explanation:

This problem is not correctly formulated. I guess that this two levels are for the hydrogen atom with the Bohr model.
If this is the case, we can apply the Rydberg formula to calculate the energy difference between two levels.

`\Delta E = hcR(1/n_f^2-1/n_i^2)=hcR(1/25-1/4)=-0.21hcR`

where `R` is the Rydberg constant, `h` is the Plank constant and `c` is the speed of the light.
The sign is negative because the Rydberg formula was derived to calculate the energy to jump from a lower to a higher state in the atom. We are calculating here the energy released by the atom. So we can simply drop the sign.

This `\Delta E` is the energy of the emitted photon that has a frequency `\nu` such that

`\Delta E=h\nu`

so we have

`h\nu=0.21hcR`

and the frequency is

`\nu=0.21\times 3\times 10^8 \times 1.097\times 10^{7}` Hz
`\nu=691.11` THz