Collection of Solved Problems in Physics

„Balmer“ series of the helium ion He⁺

Task number: 4420

• Hint 1

  What relation holds for the energy of an electron in a hydrogen‑like atom, i.e., an atom with proton number Z and a single electron? What is the energy of the ground state of the helium ion?

• Solution to hint 1

  For every hydrogen‑like atom, the relation for the energy has the same form as that for the hydrogen atom. It differs only in the proton number Z, which determines the nuclear charge.

  For the energy levels of a hydrogen‑like atom, the following relation holds \[E_n = \frac{E_1}{n^2}\,,\] where \[E_1 = -Z^2\frac{(ke^2)^2 m_e}{2\hbar^2}\] is the energy of the ground state of the atom, \(k = \frac{1}{4\pi\varepsilon_0}\) is the electrostatic constant, \(m_e\) is the electron mass, and \(e\) is the electron charge.

  We see that the ground‑state energy depends on the nuclear charge. Atoms with a higher proton number \(Z\) have a more highly charged nucleus, and the electrons are therefore more strongly attracted to it. As a result, the ground‑state energy has a larger magnitude.

  For the hydrogen atom with one proton \(Z = 1\,.\)

  Helium has a proton number of \(Z = 2\), the value of the ground‑state energy of the helium ion is therefore \[E_1^{\text{He}^+} = -2^2\frac{(ke^2)^2 m_e}{2\hbar^2} = -4\frac{(ke^2)^2 m_e}{2\hbar^2} = -4\,E_1^{\text{H}} = 4\,(-13{,}6\,\text{eV}) = -54{,}4\,\text{eV}\,,\] which is four times larger than the ground‑state energy of the hydrogen atom.

• Hint 2

  How is the wavelength of a photon emitted during an electronic transition between two states in an atom related to the energy difference corresponding to these two states?

• Solution to hint 2

  The energy levels in the hydrogen atom and the transitions between them are described in detail in the hint to problem Energy levels of the hydrogen atom.

  The energy of a photon emitted during an electronic transition from one energy level to another (denoted by the principal quantum numbers \(n_A,\,n_B\)) is given by the difference between the energies of these levels. For a hydrogen‑like atom, this difference is \[ \Delta E_{AB} = E_A - E_B = \frac{E_1}{n_A^2} - \frac{E_1}{n_B^2} = E_1\left(\frac{1}{n_A^2} - \frac{1}{n_B^2}\right) \,, \] where \(E_1\) is the energy of the ground state of the given atom.

  The energy of the emitted photon is directly proportional to its frequency and inversely proportional to its wavelength, i.e., a higher photon energy corresponds to a shorter wavelength.

  In order to determine whether the energies of photons emitted in transitions between given energy levels are greater for the hydrogen atom or for the helium ion, we must compare the energy differences between the corresponding levels of the helium ion and those of the hydrogen atom.

• Solution

  A spectral series is defined as a set of electronic transitions in an atom from higher energy levels to a level with a fixed principal quantum number n. During these transitions, a photon of a specific wavelength is emitted.

  The energy of a photon emitted during an electronic transition between two energy levels (denoted by the principal quantum numbers \(n_A\) and \(n_B\)) is given by the difference between the energies of these levels. For a hydrogen‑like atom, this difference is \[ \Delta E_{AB} = E_A - E_B = \frac{E_1}{n_A^2} - \frac{E_1}{n_B^2} = E_1\left(\frac{1}{n_A^2} - \frac{1}{n_B^2}\right) \,, \] where \(E_1\) is the energy of the ground state of the atom.

  The energy of the emitted photon is directly proportional to the photon frequency and inversely proportional to the photon wavelength. Higher energy therefore corresponds to a shorter wavelength.

  In order to determine whether the energies of photons emitted in transitions between given energy levels are greater for the hydrogen atom or for the helium ion, we must compare the energy differences between the individual levels of the helium ion and those of the hydrogen atom. For this purpose, we need to know the ground‑state energies of the helium ion and the hydrogen atom \(E_1^{\text{He}^+}\) and \(E_1^H\).

  Both hydrogen and the helium ion \(\mbox{He}^+\) are atoms consisting of a nucleus and a single electron. The number of protons in the nucleus is given by the proton number \(Z\). Such atoms are referred to as hydrogen‑like atoms.

  For any hydrogen‑like atom, the energy levels are given by \[E_n = \frac{E_1}{n^2}\,,\] where \[E_1 = -Z^2\frac{(ke^2)^2 m_e}{2\hbar^2}\] is the energy of the ground state of the atom, \(k = \frac{1}{4\pi\varepsilon_0}\) is the electrostatic constant, \(m_e\) is the electron mass, and \(e\) is the electron charge.

  We see that the ground‑state energy depends on the nuclear charge. Atoms with a higher proton number \(Z\) have a more highly charged nucleus, and the electron is therefore more strongly attracted to it. As a result, the ground‑state energy has a larger magnitude.

  For the hydrogen atom with proton number \(Z=1\), the ground‑state energy is \[E_1^{\text{H}} = -\frac{(ke^2)^2 m_e}{2\hbar^2} = -13{,}6\,\text{eV} \,. \]

  Helium has a proton number \(Z = 2\), and the ground‑state energy of the helium ion is therefore \[E_1^{\text{He}^+} = -2^2\frac{(ke^2)^2 m_e}{2\hbar^2} = -4\frac{(ke^2)^2 m_e}{2\hbar^2} = -4\,E_1^{\text{H}} = 4\,(-13{,}6\,\text{eV}) = -54{,}4\,\text{eV}\,,\] which is four times larger than the ground‑state energy of the hydrogen atom.

  The energy difference between the levels denoted by the principal quantum numbers \(n_A\) and \(n_B\) for the hydrogen atom is \[ \Delta E_{AB}^\text{H} = E_1^\text{H}\left(\frac{1}{n_A^2} - \frac{1}{n_B^2}\right) = -13{,}6\,\text{eV}\cdot\left(\frac{1}{n_A^2} - \frac{1}{n_B^2}\right) \] and for the helium ion \[ \Delta E_{AB}^{\text{He}^+} = E_1^{\text{He}^+}\left(\frac{1}{n_A^2} - \frac{1}{n_B^2}\right) = -54{,}4\,\text{eV}\cdot\left(\frac{1}{n_A^2} - \frac{1}{n_B^2}\right)\,. \]

  Thus, the energy difference between any two energy levels of the helium ion is, in absolute value, four times larger than the difference between corresponding levels of the hydrogen atom. Consequently, photons emitted in transitions in the helium ion have four times higher frequencies and shorter wavelengths than photons corresponding to the analogous transitions in the hydrogen atom. The “Balmer” series, together with the other series, is therefore shifted towards shorter wavelengths relative to the corresponding series in the hydrogen atom, i.e., towards the ultraviolet region of the spectrum.

• Answer

  The “Balmer” series is shifted towards shorter wavelengths relative to the Balmer series of the hydrogen atom, i.e., towards the ultraviolet region of the spectrum.