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)?

1 Answer
Dec 16, 2015

Answer:

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!