Collection of Solved Problems in Physics

Energy levels of the hydrogen atom

Task number: 4416

• Hint

  An electron in a hydrogen atom can occupy various states, each corresponding to a different energy. The state with the lowest energy is called the ground state, in which the electron has an energy of −13,6 eV. This state is denoted by the principal quantum number n = 1. All other states are referred to as excited states, having higher principal quantum numbers, with energies given by

  \[E_{\mathrm{n}}=\frac{E_1}{n^2} .\]

  An electron in a hydrogen atom cannot have any energy other than those given by this formula.

  If an electron transitions from one state to another, energy must either be supplied equal to the difference between the final and initial states (if the electron’s energy is to increase)

  \[E_{\mathrm{n}\rightarrow\mathrm{m}}=E_{\mathrm{m}}-E_{\mathrm{n}},\]

  or the corresponding energy is emitted in the form of a photon (a quantum of electromagnetic radiation) with this energy. The energy, frequency, and wavelength of a photon/light are related by

  \[E_{\mathrm{photon}}=hf=h\frac{c}{\lambda} .\]

• Known quantities

  c = 3,00·108 m s−1	speed of light in a vacuum
  h = 6,63·10−34 Js	Planck’s constant
  E1 = −13,6 eV	energy of the electron in a hydrogen atom in the ground state
  1 eV = 1,60·10−19 J	conversion factor between units of energy

• Solution

  a) The energy of the electron consists of potential energy and kinetic energy. Since the electron and proton attract each other, their potential energy is negative, and by convention it is set to zero at infinity (i.e., far away from the nucleus). To remove the electron from a hydrogen atom, we must supply at least enough energy to make its total energy zero. If we supply more energy, the electron will move after being “freed”, meaning that the “excess” energy will appear as the kinetic energy of the electron.

  An electron in the ground state has an energy of −13,6 eV, so to remove it we must supply 13,6 eV = 13,6·1,6·10⁻¹⁹ J = 2,2·10⁻¹⁸ J. This energy is called the ionisation energy.

  b) To excite the electron to the first excited state, we want to promote it from the state n = 1 to the state m = 2. The required energy is

  \[E_{\mathrm{excit.}}=E_{\mathrm{m}}-E_{\mathrm{n}}=\frac{E_1}{m^2}-\frac{E_1}{n^2}=\frac{E_1}{2^2}-\frac{E_1}{1^2}=-\frac{3}{4}E_1 \, ,\] \[E_{\mathrm{excit.}}=-\frac{3}{4}\cdot\left(-13{,}6\,\mathrm{eV}\right)=10{,}2\,\mathrm{eV}.\]

  The hydrogen atom must therefore absorb a photon with this energy. Its wavelength \(\lambda\) can be determined as

  \[E_{\mathrm{photon}}=E_{\mathrm{excit.}}=h\frac{c}{\lambda}\,\Rightarrow\,\lambda=\frac{hc}{E_{\mathrm{excit.}}} \, ,\] \[\lambda=\frac{6{,}63{\cdot}10^{-34}\cdot3{,}00{\cdot}10^8}{10{,}2{\cdot}1{,}60{\cdot}10^{-19}}\,\mathrm{m}=1{,}2{\cdot}10^{-7}\,\mathrm{m}=120\,\mathrm{nm}.\]

  This corresponds to an ultraviolet photon.

  c) The electric field accelerates the electron by performing work equal to the product of the electron’s charge and the voltage (i.e., the potential difference) through which it is accelerated. This work equals the energy gained by the electron

  \[W=eU=E_{\mathrm{electron}}=E_{\mathrm{excit.}},\]

  from which the required voltage is

  \[U=\frac{E_{\mathrm{excit.}}}{e}=\frac{10{,}2{\cdot}1{,}60{\cdot}10^{-19}}{1{,}60{\cdot}10^{-19}}\,\mathrm{V}=10{,}2\,\mathrm{V}.\]

  Note: The energy unit 1 eV is defined as the energy gained by an electron when accelerated through a voltage of 1 V. Therefore, to give the electron an energy of 10,2 eV, it must be accelerated through a voltage of 10,2 V.

  d) To determine which transition corresponds to this photon, we first calculate its energy

  \[E_{\mathrm{photon}}=\frac{hc}{\lambda}=\frac{6{,}63{\cdot}10^{-34}\cdot3{,}00{\cdot}10^{8}}{656{\cdot}10^{-9}}\,\mathrm{J}=3{,}03{\cdot}10^{-19}\,\mathrm{J}=1{,}89\,\mathrm{eV},\]

  and now we need to find two suitable small natural numbers such that

  \[1{,}89\,\mathrm{eV}=13{,}6\,\mathrm{eV}\cdot\left(\frac{1}{n^2}-\frac{1}{m^2}\right)\,\Rightarrow\,0{,}139=\frac{1}{n^2}-\frac{1}{m^2}.\]

  At this point, we could start testing different pairs of small natural numbers.

  However, let us think about the problem from a physical point of view and use our knowledge of the energies of the individual states.

  n	1	2	3	4	5
  |En| [eV]	13,6	3,4	1,44	0,85	0,54

  If the final state of the electron were the ground state n = 1, the electron would need to have an energy of at least (13,6 − 3,4) eV = 10,2 eV or more. Therefore, this could not be a transition to the ground state. The maximum photon energy for a transition to a given state is determined by the absolute value of the energy of the final state (this corresponds to a transition from a very highly excited state with energy close to zero). This rules out final states with n ≥ 3.

  Thus, the transition must be to the state n = 2, and simple calculation shows that the transition was from the state with m = 3. That is, the electron transitioned from the second excited state to the first excited state.

• Answer

  The energy required to remove an electron from a hydrogen atom is 13,6 eV, while its excitation to the first excited state requires 10,2 eV, corresponding to a photon with a wavelength of 120 nm or an electron accelerated through a voltage of 10,2 V. The spectral line with a wavelength of 656 nm corresponds to the transition from the second excited state to the first excited state and belongs to the so‑called Balmer series.