Question

In the Bohr model, when an electron leaves one n orbit and enters another n orbit, a photon is either emitted or absorbed. Derive a relationship between the wavelength of the emitted or absorbed photon and the change in the DeBroglie wavelength of the electron when it moves from one n to another n.

Answer 1

Bohr's Model

• Bohr began with the assumption that electrons were orbiting the nucleus, much like the earth orbits the sun.
• From classical physics, a charge traveling in a circular path should lose energy by emitting electromagnetic radiation
• If the "orbiting" electron loses energy, it should end up spiraling into the nucleus (which it does not). Therefore, classical physical laws either don't apply or are inadequate to explain the inner workings of the atom
• Bohr borrowed the idea of quantized energy from Planck
  • He proposed that only orbits of certain radii, corresponding to defined energies, are "permitted"
  • An electron orbiting in one of these "allowed" orbits:
    • Has a defined energy state
    • Will not radiate energy
    • Will not spiral into the nucleus

If the orbits of the electron are restricted, the energies that the electron can possess are likewise restricted and are defined by the equation:

Where R_H is a constant called the Rydberg constant and has the value

2.18 x 10^-18 J

'n' is an integer, called the principle quantum number and corresponds to the different allowed orbits for the electron. Thus, an electron in the first allowed orbit (closest to the nucleus) has n=1, an electron in the next allowed orbit further from the nuclei has n=2, and so on.

Thus, the relative energies of these allowed orbits for the electrons can be diagrammed as follows:

All the relative energies are negative

• The lower the energy, the more stable the atom
• The lowest energy state (n=1) is called the ground state of the atom
• When an electron is in a higher (less negative) energy orbit (i.e. n=2 or higher) the atom is said to be in an excited state
• As n becomes larger, we reach a point at which the electron is completely separated from the nucleus
  • E = (-2.18 x 10^-18 J)(1/infinity) = 0
  • Thus, the state in which the electron is separated from the nucleus is the reference or zero energy state (actually higher in energy than other states)

Bohr also assumed that the electron can change from one allowed orbit to another

• Energy must be absorbed for an electron to move to a higher state (one with a higher n value)
• Energy is emitted when the electron moves to an orbit of lower energy (one with a lower n value)
• The overall change in energy associated with "orbit jumping" is the difference in energy levels between the ending (final) and initial orbits:

DE = E_f - E_i

Substituting in for the previously defined energy equation:

When an electron "falls" from a higher orbit to a lower one the energy difference is a defined amount and results in emitted electromagnetic radiation of a defined energy (DE)

• Planck had deduced that the energy of the photons comprising EM radiation is a function of its frequency (E = h)
• Therefore, if the emitted radiation from a falling electron had a defined energy, then it must have a correspondingly defined frequency

Note:

• DE is positive when n_f is greater than n_i, this occurs when energy is absorbed and an electron moves up to a higher energy level (i.e. orbit).
• When DE is negative, radiant energy is emitted and an electron has fallen down to a lower energy state

Revisiting Balmer's equation:

In 1885 a Swiss school teacher figured out that the frequencies of the light corresponding to these wavelengths fit a relatively simple mathematical formula:

where C = 3.29 x 10¹⁵ s^-1 (not the 'c' used for the speed of light)

Since energy lost by the electrons is energy "gained" by the emitted EM energy, the EM energy from Bohr's equation would be:

Thus, Balmer's constant 'C' = (R_H/h) (Rydberg constant divided by Planck's constant), and n_f = 2. Thus, the only emitted energies which fall in the visible spectrum are from those electrons which fell down to the second quantum orbital. Those which fell down to the first orbital have a higher energy (frequency) than can be seen in the visible spectrum.

Calculate the wavelength of light that corresponds to the transition of the electron from the n=4 to the n=2 state of the hydrogen atom. Is the light absorbed or emitted by the atom?

Since the electron is "falling" from level 4 down to level 2, energy will be given up and manifested as emitted electromagnetic radiation:

DE = (2.18 x 10^-18 J)((1/16)-(1/4)) = -4.09 x 10^-19 J (light is emitted)

4.09 x 10-19 J = (6.63 x 10^-34 Js) * (n)

6.17 x 10¹⁴ s^-1 = n

l = (3.00 x 10⁸ m s^-1)/ (6.17 x 10¹⁴ s^-1) = 4.87 x 10^-7m = 487 nm

Bohr's model of the atom was important because it introduced quantized energy states for the electrons. However, as a model it was only useful for predicting the behavior of atoms with a single electron (H, He⁺, and Li²⁺ ions). Thus, a different model of the atom eventually replaced Bohr's model. However, we will retain the concept of quantized energy states