Question

Question #10f96

Answer 1

The wavelength is `lambda=1.28xx10^(-6)m`

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=3`, then subtract these values:

`E_3=(-1311)/3^2= 145.67 (kJ)/(mol)`

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

Subtract and you get `DeltaE=93.23 (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=93.23/(3.99xx10^(-13))=2.34xx10^(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)/(2.34xx10^14)= 1.28xx10^(-6)m`

which is in the infrared portion of the visible spectrum.