Question

Using Rydberg's equation for the energy levels (E = -R/n^2) and the wavelength and the c=(lambda)v, E=hv equations, SHOW/ PROVE how E(red band in H-spectrum, lambda = 656 nm) = (Delta or change/ difference) E (3 --> 2)?

Answer 1

This is a bit long and probably there is a faster way but I tried this:

Explanation:

Ok; you know that when an electron jumps from an allowed orbit to another it absorbs/emits energy in form of a photon of energy `E=hnu` (where `h=`Planck's Constant and `nu=`frequency).

Your "red" photon represents a transition between two orbits (of quantum numbers `n` and `n+1`) separated by a "energy" of:

`E_(red)=h*nu_(red)`

but red means a wavelength: `lambda_(red)=656nm`

so the "red" frequency will be: `nu_(red)=c/lambda_(red)=(3xx10^8)/(656xx10^-9)=4.57xx10^14Hz`

And energy:

`E_(red)=6.63xx10^-34*4.57xx10^14=3xx10^-19J=1.87eV`
(where `1eV=1.6xx10^-19J`)

Now we need to find the two orbits (their quantum numbers) from where and toward where the electron jumped:
We use Rydberg's Formula where we know that:
`DeltaE=E_f-E_i=1.87eV`
and also:
`DeltaE=-R(1/(n+1)^2-1/n^2)`
`1.87=-13.6(1/3^2-1/2^2)`
(I used Rydberg Constant in `eV`; `R=13.6eV`)
I found:
`1.87=1.88` that works fine I think!

Hope it helps!