Wave Mechanical Model of the Hydrogen Atom
- With Bohr's quantum hypothesis, it became possible to combine classical mechanics with quantum wave mechanics, to produce a satisfactory model of atomic structure.
- This model allows us to:
- explain existing observations.
- predict new observation.
- Given the proton has a mass of mp and charge of e+ , and the electron has a mass of me and charge of e-, with radius of orbit rn, and velocity vn, then if n equals the number of de Broglie wavelengths in the nth orbit, we can write:
Fc = Fe
(me(vn)2)/rn = (ke2)/(rn)2 equation 1
from 2prn = (nh)/(mevn) we can get vn = (nh)/(2pmern) equation 2
- Substituting vn (that is equation 2) into equation 1 above and rearranging gives:
rn = (n2h2)/(4p2meke2) equation 3
- rn = nth radius of the hydrogen atom.
- By substituting in values for h, p , me, k, and e gives:
rn = (5.3 X 10-11m)n2
- For the smallest orbit, n = 1 and r1 = 5.3 X 10-11m (called the Bohr radius).
- The Bohr radius matches closely the value of the measured size of the hydrogen atom.
- By substituting equation 3 into the equation for vn (equation 2) we find:
vn = (nh)/(2pme((n2h2)/(4p2kmee2)))
- Rearranging this expression gives:
vn = (2pke2)/(nh) equation 4
- Substituting in p , k, e, and h gives:
vn = (2.2 X 106m/s)/n
- This expression allows us to calculate the velocity of the electron in its various orbits.
- The energy in the nth orbit would be the sum of the potential energy and the kinetic energy.
En = Ep + Ek
En = ((1/2)me(vn)2) + ((-ke2)/rn) equation 5
- Substituting the general expressions for the radius and speed of the electron (equation 3 and equation 4) into equation 5 gives:
En = (1/2)me((2pke2)/(nh))2 - (ke2)/((n2h2)/(4p2kmee2))
- Expanding and rearranging gives:
En = (-2p2mek2e4)/(n2h2)
- Substituting for the constants gives:
En = (-2.17 X 10-18J)/n2 = (-13.6 eV)/n2
Therefore E1 = -2.17 X 10-18J = -13.6 eV
- Remember that En = -Rhc/n2. Substituting information for the first energy level into this expression allows us to solve for Rydberg's constant.
-2.17 X 10-18J = -R(6.63 X 10-34Js)(3.00 X 108m/s)/(1)2
or R = 1.09 X 107m-1
- This calculation of Rydberg's constant gave Bohr confidence enough to publish his work.
- This wave - mechanical model for the hydrogen atom matches exactly the evidence from the atom's emission spectrum. We can calculate the energies for the various stationary states, and then calculate:
DE = E2 - E1
and compare to the emission spectrum.
- We can now draw a complete energy level diagram for hydrogen.