Question

How do we work out the frequency of transition for an electronic transition of the hydrogen ATOM, whose electron jumps from `n=2` to `n=5`?

Answer 1

Just to retire this question, we must presume that we examine the transitions of an hydrogen atom.....this was not made clear in the question.

Explanation:

And for this process we use the Rydberg formula,

`1/lambda_("vacuum")=R{1/(1/n_1^2-1/n_2^2)}`,.........

......where `R="the Rydberg constant,"` `1.097 xx 10^7 *m^(−1);`

`n_1=2, n_2=5`........

`1/lambda_("vacuum")=1.097xx10^7*m^-1{1/(1/2^2-1/5^2)}`

`1/lambda_("vacuum")=1.097xx10^7*m^-1{1/(1/4-1/25)}=52238095.2*m^-1`

And thus `lambda=8.90xx10^-8*m`, if I have done my arithmetic correctly, and I would check it!

Now since we deal with the electromagnetic spectrum,

`c=nuxxlambda`, and thus ................

`c/lambda=nu=(3.00xx10^8*m*s^-1)/(8.90xx10^-8*m)=3.371xx10^15*s^-1`

And note that we get an answer in `s^-1`, which is appropriate for a frequency.

So we gots `"wavelength"`, and `"frequency"`, we need to find the energy in `kJ*mol^-1`. We use the old Planck relationship, i.e. `epsilon=hnu`, where `h=6.626xx10^(-34)*J *s=2.23xx10^-18*J` for a SINGLE PHOTON.

Bu we gots a mole of photons......and thus...

`epsilon_(mol^-1)=2.23xx10^-18*Jxx6.022xx10^23*mol^-1`

`=1.35xx10^6*J*mol^-1=1.35xx10^3*kJ*mol^-1`.

Again you should check my calculations: `"all care taken, but no responsibility admitted."`