Rydberg plus Coulomb equals Bohr.
From your Rydberg lab, you have the relation for the wavelength
of light emitted for electrons in the excited states above n=2:
for n = 3, 4, 5,...
We can derive a similar expression using Bohr's approach with
classical electric fields.
a) First, the Coulomb force is the centripetal force; set the
two expressions equal to each other and solve for mv.
b) Secondly, the angular momentum of the orbiting electron is
quantized. Set the expression for the classical angular
momentum equal to nħ and solve the resulting equation for r.
c) For n=1, determine the orbital radius r.
q = e = -1.6 x 10^-19 [C]
ħ = 1.055 x 10^-34 [Js]
ke = 9 x 10^9 [Nm^2/C^2]
me = 9.09 x 10^-31 [kg]
d) Show that the kinetic energy = -(1/2) U, where U = the
electric potential energy (use the approach from part a).
Then use your expression (not the value) for r in part b to write
the remaining (1/2)U in terms of 1/n^2.
e) Calculate a value for the coefficient of 1/n^2 in
the expression from the previous part.
f) That expression is the ionization energy in terms of the
quantum number n. Write the difference in energy levels (the energy
of the photon emitted) in terms of h, c, and λ, set it equal to the
expression in e) and rearrange to get the same form as Rydberg's
equation.
g) Calculate the coefficient in front of the difference in 1/λ
and compare it to Rydberg's constant.