Collection of Solved Problems in Physics

Energy spectrum of positronium

Task number: 4419

• Theory – reduced mass

  Positronium is a hydrogen‑like atom – it consists of a positively charged nucleus and a single electron (with the difference that instead of a proton, the nucleus is formed by a positron). Therefore, the expression for the electron’s energy in positronium is similar to that of the hydrogen atom. The energy of the \(n\)‑th energy level is \[E_n = \frac{E_1}{n^2}\,,\] where \(E_1\) is the ground‑state energy.

  In deriving the energy expression, the electron‑nucleus system is treated as a two‑body problem, similar to problems of two point masses in classical mechanics. Instead of considering the motion of the electron and the nucleus relative to the surroundings, we separately handle the motion of their centre of mass (in an atom with very heavy nucleus, the centre of mass is essentially at the nucleus) and the relative motion of the electron with respect to the nucleus (for a very heavy nucleus, this relative motion can be identified with the electron’s motion relative to the surroundings). As in classical mechanics, we replace the electron mass \(m_e\) with the so‑called reduced mass \[\mu = \frac{m_e m_N}{m_e + m_N}\] and the nucleus mass \(m_N\) with the total mass of the atom \[M = m_e + m_N \,.\] The ground‑state energy is then guven by \[E_1 = -\frac{(ke^2)^2 \mu}{2\hbar^2} \,.\]

• Hint a)

  Compare the reduced mass with the electron mass

  i) in case of the hydrogen atom,

  ii) in case of positronium.

  Is the relative difference between the reduced mass and the electron mass in both cases comparable to the precision required for determining the energy levels?

• Solution a) i)

  In the case of the hydrogen atom, the nuclear mass \(m_N\) is approximately two thousand times greater than the electron mass, \(m_N = 2000 \,m_e\).

  The reduced mass for the hydrogen atom \(\mu_H\) is calculated as follows \[\mu_H = \frac{m_N m_e}{m_N + m_e} = \frac{2000 \,m_e^2}{2001 \,m_e} \,\dot{=}\, 0{,}9995\,m_e \,. \]

  The difference between the reduced mass and the electron mass is \[\Delta\mu_H = m_e - \mu_H = m_e - 0{,}9995\,m_e = 0{,}0005\,m_e \,,\] and their relative difference is \[\frac{\Delta\mu_H}{m_e} = 0{,}0005 \,.\] Thus, these two masses differ by a fraction of a per mille.

  This difference can be neglected in a first approximation, and when determining the energy levels of the hydrogen atom, one can consider only the electron mass instead of the reduced mass.

• Solution a) ii)

  In the case of positronium, the mass of the “nucleus” and the electron are the same, \(m_N = m_e\).

  The reduced mass for positronium \(\mu_P\) is calculated as \[\mu_P = \frac{m_e m_e}{m_e + m_e} = \frac{m_e^2}{2m_e} = \frac{m_e}{2} \,. \]

  The difference between the reduced mass and the electron mass is \[\Delta\mu_P = m_e - \mu_P = m_e - \frac{m_e}{2} = \frac{m_e}{2} \,,\] so the electron mass is twice as large as the reduced mass.

  This difference is no longer negligible, so when determining the energy levels of positronium, the reduced mass must be used rather than just the electron mass.

• Solution b)

  In the previous section, we showed that when determining the energy levels of positronium, we must use the reduced mass instead of the electron mass, because in the case of positronium the mass of the “nucleus” is not much greater than the electron mass, unlike in the hydrogen atom.

  The ground‑state energy of positronium is given by \[E_1^P = -\frac{(ke^2)^2 \mu_P}{2\hbar^2} = -\frac{1}{2}\frac{(ke^2)^2 m_e}{2\hbar^2}\,,\] where \(\mu_P = \frac{m_e}{2}\) is the reduced mass of positronium.

  Note that the expression \[-\frac{(ke^2)^2 m_e}{2\hbar^2}\] is the ground‑state energy of the hydrogen atom \[E_1^H = -13{,}6\,\text{eV} \,.\] For the hydrogen atom, the electron mass \(m_e\) can be used instead of the reduced mass with sufficient accuracy. Thus, the ground‑state energy of positronium can be calculated as \[ E_1^P = \frac{E_1^H}{2} = \frac{-13{,}6\,\text{eV}}{2} = -6{,}8\,\text{eV} \,. \]

• Hint c)

  The transitions between energy levels of an atom and their relation to the atomic spectrum are described in Hint 2 of problem „Balmer“ series of the helium ion He⁺.

• Solution c)

  The energy of a photon emitted during a transition of an electron in positronium from one energy level to another (denoted by the principal quantum numbers \(n_A,\,n_B\)) is \[ \Delta E_{AB} = E_A - E_B = \frac{E_1^P}{n_A^2} - \frac{E_1^P}{n_B^2} = E_1^P\left(\frac{1}{n_A^2} - \frac{1}{n_B^2}\right)\,, \] where \(E_1^P\) is the ground‑state energy of positronium.

  The photon emitted during the transition of positronium from the first excited state with \(n_B = 2\) to the ground state with \(n_A = 1\) has the energy \[\Delta E_{AB} = E_1^P\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = \frac{3}{4}\,E_1^P = \frac{3}{4} \cdot 6{,}8\,\text{eV}= 5{,}1\,\text{eV}\,.\]

  The relationship between the energy of the emitted photon, its frequency, wavelength, and the speed of light is \[\Delta E_{AB} = hf = h\frac{c}{\lambda} \,.\] The wavelength of the photon emitted in the transition from the first excited state to the ground state of positronium is therefore \[\lambda = h\frac{c}{\Delta E_{AB}} \,. \]

  By substituting numerical values and converting the energy from electronvolts to SI units, we obtain the wavelength \[ \lambda = 6{,}626\,\cdot\,10^{-34} \cdot \frac{3\,\cdot\,10^8}{(5{,}1\,\cdot\,1{,}602\,\cdot\,10^{-19})} \,\dot{=}\, 243\,\text{nm} \,. \]

  We see that the wavelength of the transition from the first excited state to the ground state of positronium falls within the ultraviolet region of the spectrum.

• Solution d)

  The energy difference between the first excited state and the ground state of the hydrogen atom is \[\Delta E_{AB} = E_1^H\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = \frac{3}{4}\,E_1^H = \frac{3}{4} \cdot 13{,}6\,\text{eV}= 10{,}2\,\text{eV}\,,\] which is therefore twice as large as in the case of positronium.

  The photon emitted in this transition thus has twice the energy of the corresponding transition in positronium.

  The relationship between the energy of the emitted photon, its frequency, wavelength, and the speed of light is \[\Delta E_{AB} = hf = h\frac{c}{\lambda} \,.\] The wavelength of the photon emitted in the transition from the first excited state to the ground state of the hydrogen atom is therefore \[\lambda = h\frac{c}{\Delta E_{AB}} \,. \]

  By substituting numerical values and converting the energy from electronvolts to SI units, we obtain the wavelength \[ \lambda = 6{,}626\,\cdot\,10^{-34} \cdot \frac{3\,\cdot\,10^8}{(10{,}2\,\cdot\,1{,}602\,\cdot\,10^{-19})} \,\dot{=}\, 122 \,\text{nm} \,. \]

  The wavelength of the transition from the first excited state to the ground state of the hydrogen atom in the Lyman series also falls within the ultraviolet region of the spectrum and is even shorter than the corresponding transition in positronium.

• Answer

  a) In the case of positronium, unlike hydrogen, it is necessary to consider the reduced mass instead of the electron mass, because the “nucleus” of positronium and the electron have the same mass. We cannot treat the “nucleus” as infinitely heavy, as in hydrogen, where the nucleus is about two thousand times heavier than the electron.

  The relative difference between the electron mass and the reduced mass is on the order of tenths of a per mille for hydrogen, whereas for positronium it is 50 %.

  b) The ground‑state energy of positronium is \(6{,}8\,\text{eV}\).

  c) The wavelength of the photon emitted in the transition of positronium from the first excited state to the ground state is \(243\,\text{nm}\).

  d) The wavelength of the photon emitted in the corresponding transition in the hydrogen atom in the Lyman series is \(122\,\text{nm}\), i.e., half the wavelength of the positronium transition.