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7 Quantum theory
7A The origins of quantummechanics
Answers to discussion question
D7A.2 �e ultimately unsuccessful classical approach to the description of black-body
radiation involved assuming that the radiation resulted fromoscillating electric
charges in the walls of the body, and that each oscillator has the same average
energy as predicted by the equipartition principle. �is view results in the
ultra-violet catastrophe, in which the radiation increases without limit as the
wavelength becomes shorter.
Planck assumed two things: �rst, that the oscillators could only have energies
given by E = nhν, where ν is the frequency and n is 0, 1, 2, . . .; second, that the
probability of an individual oscillator having a particular energy is described
by the Boltzmann distribution. As the frequency of the oscillator or the value
of n increases, so does its energy and the Boltzmann distribution predicts that
such a state is less likely. In addition, the highest frequency oscillations may
not be excited at all, that is have n = 0, on the grounds that the resulting state
has too high an energy to be populated. Planck’s theory therefore avoids the
ultraviolet catastrophe by having no excitation of highest frequency oscillators.
D7A.4 By wave-particle duality it is meant that in some experiments an entity be-
haves as a wave while in other experiments the same entity behaves as a par-
ticle. Electromagnetic radiation behaves as a wave in di�raction experiments
but it behaves as particulate photons in absorption and emission spectroscopy.
Electrons behave as waves in di�raction experiments, but as particles in the
photoelectric e�ect.
�e development of quantum theory is much concerned with the need to em-
brace this wave-particle duality and, as is explained in the following Topics,
this is exempli�ed by the introduction of the wavefunction to describe the
properties of a particles and the notion of ‘complementary variables’ such as
position and momentum.
Solutions to exercises
E7A.1(b) Wien’s law [7A.1–238], λmaxT = 2.9 × 10−3 mK, is rearranged to give the
wavelength at which intensity is maximised
λmax = (2.9 × 10−3 mK)/T = (2.9 × 10−3 mK)/(2.7 K) = 1.1 × 10−3 m