Question

How do you calculate the frequency and wave length of hydrogen atom spectrum when electron transfer from 5 = 5 to n = 2?

Answer 1

The frequency is `6.9xx10^(14) Hz` and the wavelength is `4.35xx10^(-7) m`
The calculations used to find these values are shown below...

Explanation:

To answer this question, we start with Bohr's result for the energies of the stationary states of hydrogen. These are the quantized, allowable energies the electrons may possess.

The equation is

`E_n=-((e^4m)/(8epsilon_o^2h^2))(1/n^2)`

Depending on the units you use, the first bracket is a collection of constants, and can be replaced with a single value.

In Chemistry-friendly terms, I will set this value to be `1311 (kJ)/(mol)`

Thus, `E_n=(-1311)/n^2` gives the allowed energy levels.

Now, we evaluate the energy for `n=5` and again for `n=2`, then subtract these values:

`E_2=(-1311)/2^2= 327.75 (kJ)/(mol)`

`E_5=(-1311)/5^2= 52.44 (kJ)/(mol)`

Subtract and you get `DeltaE=275.31 (kJ)/(mol)`

This change in the energy of the atom equals the energy carried off by the photon that is released.

To convert to frequency, we apply Planck's relation:

`E=hf` where `h=3.99xx10^(-13) (kJs)/(mol)` is Planck's constant, in units consistent with our earlier choice.

`f=(DeltaE)/h=275.31/(3.99xx10^(-13))=6.9xx10^(14) Hz`

as the frequency.

The wavelength comes from the relation `v=flambda` where `v` is the speed of light, `3.0 xx 10^8 m/s`)

Finally `lambda=(3.0xx10^8)/(6.9xx10^14)= 4.35xx10^(-7)m`

which is in the violet portion of the visible spectrum.