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White dwarfs and neutron stars
White dwarfs and neutron stars

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3.1 The de Broglie wavelength of electrons and nucleons

Let’s begin by thinking in terms of the quantum mechanical properties of particles. A volume of space containing a gas may be considered to be occupied by particles whose wave properties give rise to standing waves that fit neatly within that volume. They therefore have quantised values of wavelength, and also of energy and momentum. This quantisation means that there is a discrete (rather than continuous) distribution of possible particle energies.

The number of possible quantum states available to particles below some energy is finite; often it is very large, but nevertheless it is finite. In the very dense interiors of some astronomical objects, such a finite number of available quantum states, though large, can be insufficient for the huge number of particles squeezed into the confined volume. Under these circumstances, physics then takes new twists, giving rise to some remarkable properties.

The critical point of having too few available quantum states for the number of particles present is reached when the separations of the gas particles become smaller than their de Broglie wavelength. At that point, the wave properties of the gas dominate its classical properties, and a quantum gas is said to exist. The following activity explores how the de Broglie wavelengths for protons and electrons compare in the core of the Sun.

Activity 1

  • a.The de Broglie wavelength of non-relativistic particles is given by Equation (2) as lamda sub dB equals h solidus left parenthesis three times m times k sub cap b times cap t right parenthesis super one solidus two where m is the particle’s mass, kB is Boltzmann’s constant ( equals 1.38 multiplication 10 super negative 23 cap j cap k super negative one ) and T is the temperature. Calculate the de Broglie wavelength, in the core of the Sun at a temperature of cap t sub c comma circled dot operator equals 15.6 multiplication 10 super six cap k , of:

    • i. a proton (with mass m sub p equals 1.67 multiplication 10 super negative 27 kg)
    • ii.an electron (with mass m sub e equals 9.11 multiplication 10 super negative 31 kg)
  • b.Which has the larger de Broglie wavelength, and by what factor?

  • c.How does this ratio vary from star to star?

Comment

  • a.(i) For a proton in the core of the Sun:

    multiline equation row 1 lamda sub dB of p equals h solidus left parenthesis three times m sub p times k sub cap b times cap t sub c comma circled dot operator right parenthesis super one solidus two row 2 Blank equals 6.63 multiplication 10 super negative 34 cap j s divided by Square root of three multiplication 1.67 multiplication 10 super negative 27 kg prefix multiplication of 1.38 multiplication 10 super negative 23 cap j cap k super negative one multiplication 15.6 multiplication 10 super six cap k row 3 Blank equals 6.38 multiplication 10 super negative 13 m full stop

    (ii) For an electron in the core of the Sun:

    multiline equation row 1 lamda sub dB of e equals h solidus left parenthesis three times m sub e times k sub cap b times cap t sub c comma circled dot operator right parenthesis super one solidus two row 2 Blank equals 6.63 multiplication 10 super negative 34 cap j s divided by Square root of three multiplication 9.11 multiplication 10 super negative 31 kg prefix multiplication of 1.38 multiplication 10 super negative 23 cap j cap k super negative one multiplication 15.6 multiplication 10 super six cap k row 3 Blank equals 2.73 multiplication 10 super negative 11 m full stop
  • b.The ratio of their de Broglie wavelengths is therefore:

    multiline equation row 1 lamda sub dB of e divided by lamda sub dB of p equation sequence part 1 equals part 2 h divided by left parenthesis three times m sub e times k sub cap b times cap t sub c comma circled dot operator right parenthesis super one solidus two multiplication left parenthesis three times m sub p times k sub cap b times cap t sub c comma circled dot operator right parenthesis super one solidus two divided by h equals part 3 left parenthesis m sub p divided by m sub e right parenthesis super one solidus two row 2 Blank equation sequence part 1 equals part 2 left parenthesis 1.67 multiplication 10 super negative 27 kg divided by 9.11 multiplication 10 super negative 31 kg right parenthesis super one solidus two equals part 3 42.8

    i.e. the de Broglie wavelength of the electron is greater than that of the proton, by a factor of about 40.

  • c.From part (b), this ratio depends only on the mass of the particles, and hence is independent of the environment and hence of the temperature. The electron’s wavelength is about 40 times longer than the proton’s wavelength in all stars.

Note that neutrons have a similar de Broglie wavelength to protons because both types of particle have similar mass. Since the de Broglie wavelength of electrons is roughly 40 times longer than that of nucleons (i.e. protons or neutrons), then as the stellar core contracts and its density increases, electrons run out of space and hence reach the quantum limit corresponding to degeneracy sooner than nucleons.