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Recall from last lecture:

28.3 The Bohr Theory of Hydrogen

I will repeat and paraphrase the four assumptions of the Bohr theory:

  1. The electron moves in a circle about the proton under the influence of the Coulomb force.
  2. Only certain orbits are allowed (meta-stable).
  3. An electron can jump between allowed orbits by emitting or absorbing a photon with the proper energy given by: Ei - Ef = hf.
  4. The allowed orbits satisfy the angular momentum requirement:
    mev r = n hbar
    where n = 1, 2, 3, ..., and hbar = h/2p.

From these assumptions we can solve for the allowed orbits and energy levels of the hydrogen atom. We begin with a picture of the electron in a circular orbit around the proton (Figure 28.5). The electric potential energy of the atom is

PE = ke q1 q2 / r = -ke e2 / r

The total energy of the atom is the sum of the kinetic energy of the electron (the proton is at rest) and potential energy

E = KE + PE = (1/2)me v2 - ke e2 / r
(Notice that the hydrogen at is non-relativisitic, so we neglect the energy associated with the mass.)

For an object in circular orbit, the centripetal force must equal mv2/r. The centripetal force on the electron is the Coulomb force:

ke e2 / r2 = me v2 / r
If we now multiply both sides by r and divide by 2, we see that:
ke e2 / 2r = (1/2) me v2
The left hand side is -1/2 the PE and the right hand side is the KE. Substituting this into the expression for the energy, we get:
E = -ke e2 / 2r

From the condition for allowed orbits, we get:

vn = n hbar / me rn
where I've added the subscript n to remind us that this value depends on the energy level of the electron. If we square this, we can compare to v2 from the centripetal force equation:
vn2 = n2 hbar2 / me2 rn2 = ke e2 / me rn

From the above expression, we can get an expression for the radii

rn = n2 hbar2 / me ke e2
where n can take on the integer values 1, 2, 3, ... The radius of the smallest orbit (n=1) is called the Bohr radius, a0
a0 = hbar2 / me ke e2 = 0.0529 nm
We can express the rn in terms of a0 as
rn = n2 a0

Now, substituting the expression for the radii into the energy expression we get:

En = -(me ke2 e4 / 2 hbar2) (1/n2)
This is our sought after result! If we substitute for the constants, we can write that the energy of the nth Bohr orbit is
En = -13.6eV / n2
The lowest energy state (n=1) has energy E1 = -13.6eV. The energy is negative because the electron is bound to the proton, and we used the expression for potential energy which is zero when the electron and proton are widely separated. This situation corresponds to n=infinity.

When an electron jumps between energy levels, the atom can emit or absorb a photon. Let's concentrate on emission, for absorption just reverse the initial and final state. In emission, the initial state is just the atom in its initial energy level, and the final state is the atom in its final energy level plus a photon. Conservation of energy give us:

Ei = Ef + hf
so that hf = Ei - Ef.

Supposing that the inital energy level is ni and the final level is nf, then the wavelength of the emitted photon is:

1/l = f/c = (Ei - Ef) / hc = (me ke2 e4 / 2hc hbar2) {(1/nf2) - (1/ni2)}
Replace h with 2p hbar, and the constant piece can be equated with Rydberg's constant:
RH = me ke2 e4 / 4p c hbar3
The agreement between the value of RH determined from measurements of the Balmer series, and this expression are remarkably good. That agreement was recognized as a major accomplishment of Bohr's theory.

Draw energy levels for hydrogen and show transitions corresponding to Balmer and Lyman series (Figure 28.7).

Example: P28.9

Show that the speed of the electron in the nth Bohr orbit in hydrogen is given by vn = kee2 / n hbar.

From the last of Bohr's assumptions we have
mevn r = n hbar    (1)
and equating the Coulomb force with the centripetal force gives
mevn2 / r = kee2 / r2.    (2)
If we multiply (2) by r2 we get
mevn2 r = kee2.    (3)
Now divide (3) by (1) and we get
vn = kee2 / n hbar.

Example: P28.12

Four possible transitions for a hydrogen atom are listed below:
(A) ni = 2; nf = 5 (B) ni = 5; nf = 3
(C) ni = 7; nf = 4 (D) ni = 4; nf = 7
(a) Which transition will emit the shortest wavelength photon? (b) For which transition will the atom gain the most energy?

(a) First, to emit a photon, the initial level must be greater than the final level. This is only true for transitions (B) and (C). Transitions (A) and (D) require absorption of a photon. Second, the shortest wavelength photon comes from the transition with the largest energy difference. Using En = -(13.6eV / n2), and Ei = Ef + hf for emission, we find that for transition (B)
Ei - Ef = -13.6eV (1/5² - 1/3²) = 0.967eV
and for transition (c)
Ei - Ef = -13.6eV (1/7² - 1/4²) = 0.572eV.
Therefore transition (B) will emit the shortest wavelength photon.

(b) The atom will gain energy when it absorbs a photon, Ei + hf = Ef. For transition (A), the change in energy is
Ef - Ei = -13.6eV (1/5² - 1/2²) = 2.86eV
and for transition (D)
Ef - Ei = -13.6eV (1/7² - 1/4²) = 0.572eV.
Therefore, the atom in transition (A) gains the most energy.

Example: P28.22 (optional)

Two hydrogen atoms, both initially in the ground state, undergo a head-on collision. If both atoms are to be excited to the n=2 level in this collision, what is the minimum speed each atom can have before the collision?

Recall that collisions can be elastic (momentum and kinetic energy are conserved) or inelastic (momentum is conserved, but not kinetic energy). In this problem, we want to convert some of the kinetic energy of the atoms into internal energy (exciting the electrons to the n=2 level), therefore this is an inelastic collision. If kinetic energy is converted to internal energy as efficiently as possible, then this is a totally inelastic collision -- the two atoms will be stuck together afterwards. Momentum is conserved if the atoms have the equal and opposite initial momenta (velocities), and, after colliding, are at rest.

Energy is conserved if the initial energy (kinetic energy plus ground state energy) equals the final energy (excited state energy)
2E1 + 2K.E. = 2E2
Divide out the common factor of 2 such that
-13.6eV / 1² + (1/2)mHv2 = -13.6eV / 2²
where mH is the mass of a hydrogen atom, which is approximately the same as the mass of a proton, which is 938MeV / c². Solve for v
v2 = 13.6eV (1 - 1/4) 2 / mH = 20.4eV / 938MeV/c² = 2.2×10-8c²
v = 1.5×10-4c = 4.5×104m/s.

© Robert Harr 2000