IChO 2001 · Theory problem 1

Hydrogen Atom and Hydrogen Molecule

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Figure 1

Niels Bohr (1885-1962)

The observed wavelengths in the line spectrum of hydrogen atom were first expressed in terms of a series by Johann Jakob Balmer, a Swiss teacher. Balmer’s empirical formula is

\[ \frac{1}{\lambda} = R_{\mathrm H}\left(\frac{1}{2^2}-\frac{1}{n^2}\right); \qquad n=3,4,5,\ldots \]

Here,

\[ R_{\mathrm H} = \frac{m_e e^4}{8\varepsilon_0^2h^3c} = 109\,678\ \mathrm{cm^{-1}} \]

is the Rydberg konstant, \(m_e\) is the mass of an electron. Niels Bohr derived this expression theoretically in 1913. The formula is easily generalized to any one electron atom/ion.

1.1 Calculate the longest wavelength in Å (\(1\ \text{Å}=10^{-10}\ \mathrm m\)) in the ‘Balmer series’ of singly ionized helium (\(\ce{He+}\)). Ignore nuclear motion in your calculation.

1.2 A formula analogous to Balmer’s formula applies to the series of spectral lines which arise from transitions from higher energy levels to the lowest energy level of hydrogen atom. Write this formula and use it to determine the ground state energy of a hydrogen atom in eV.

A ‘muonic hydrogen atom’ is like a hydrogen atom in which the electron is replaced by a heavier particle, the muon. The mass of a muon is about 207 times the mass of an electron, while its charge is the same as that of an electron. A muon has a very short lifetime, but we ignore its unstable nature here.

1.3 Determine the lowest energy and the radius of the first Bohr orbit of the muonic hydrogen atom. Ignore the motion of the nucleus in your calculation. The radius of the first Bohr orbit of a hydrogen atom (called the Bohr radius, \(a_0=\dfrac{\varepsilon_0h^2}{m_e e^2\pi}\)) is \(0.53\ \text{Å}\).

The classical picture of an "orbit" in Bohr’s theory has now been replaced by the quantum mechanical notion of an ‘orbital’. The orbital \(\psi1\sigma_{1s}(r)\) for the ground state of a hydrogen atom is given by

\[ \Psi_{1s}(r) = \frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0} \]

where \(r\) is the distance of the electron from the nucleus and \(a_0\) is the Bohr radius.

1.4 Consider a spherical shell of radius \(a_0\) and thickness \(0.001a_0\). Estimate the probability of finding the electron in this shell. Volume of a spherical shell of inner radius \(r\) and small thickness \(\Delta r\) equals \(4\pi r^2\Delta r\).

The \(\ce{H2}\) molecule can dissociate through two different channels:

Figure 2

The graph of energy (\(E\)) vs internuclear distance (\(R\)) for \(\ce{H2}\) is shown schematically in the figure. The atomic and molecular energies are given in the same scale.

1.5 Put appropriate channel labels (i) or (ii) in the boxes below.

1.6 Determine the values of the dissociation energies (\(D_e\) in eV) of the \(\ce{H2}\) molecule corresponding to

channel (i) \(\boxed{\phantom{000000}}\)

channel (ii) \(\boxed{\phantom{000000}}\)

Figure 3

1.7 From the given data, calculate the energy change for the process

Figure 4

1.8 \(\ce{H-}\) is a two-electron atomic system. Assuming that the Bohr energy formula is valid for each electron with nuclear charge \(Z\) replaced by \(Z_{\mathrm{eff}}\), calculate \(Z_{\mathrm{eff}}\) for \(\ce{H-}\).

Official English text of the problem.

Answers

  • 1.1 \(\boxed{\lambda_{\max}=1.641\times10^3\ \text{Å}}\), for the \(n=3\rightarrow2\) transition.
  • 1.2 \(\boxed{\lambda^{-1}=R_{\mathrm H}(1-n^{-2}),\quad n=2,3,\ldots}\); \(\boxed{E_1=-13.6\ \mathrm{eV}}\).
  • 1.3 \(\boxed{E_1^{(\mu)}\approx-2.81\times10^3\ \mathrm{eV}}\); \(\boxed{r_1^{(\mu)}\approx2.6\times10^{-3}\ \text{Å}=2.6\times10^{-13}\ \mathrm m}\).
  • 1.4 \(\boxed{P\approx5.41\times10^{-4}=0.0541\%}\).
  • 1.5 Upper box: (ii). Lower box: (i).
  • 1.6 Channel (i): \(\boxed{D_e=4.7\ \mathrm{eV}}\). Channel (ii): \(\boxed{D_e=17.6\ \mathrm{eV}}\).
  • 1.7 \(\boxed{\Delta E=+0.7\ \mathrm{eV}}\); energy must be supplied.
  • 1.8 \(\boxed{Z_{\mathrm{eff}}=0.725}\).

Solution

The species used below are the hydrogen atom \(\ce{H}\), [H]; hydrogen molecule \(\ce{H2}\), [H][H]; singly ionized helium \(\ce{He+}\), [He+]; proton \(\ce{H+}\), [H+]; and hydride ion \(\ce{H-}\), [H-]. Ordinary SMILES cannot represent a free electron, replacement of an electron by a muon, or an orbital wavefunction.

Energies are measured relative to separated nuclei and free electrons at rest. For numerical conversion, \(hc=1.23984198\times10^{-4}\ \mathrm{eV\,cm}\).

1.1 — Longest wavelength in the helium-ion Balmer series

For a one-electron ion of nuclear charge \(Z\), neglecting nuclear motion,

\[ \frac{1}{\lambda} = R_{\mathrm H}Z^2 \left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right). \]

The Balmer series has \(n_f=2\). Its longest wavelength corresponds to its smallest transition energy, hence \(n_i=3\). For singly ionized helium, \(Z=2\):

\[ \frac{1}{\lambda_{\max}} = 4R_{\mathrm H}\left(\frac14-\frac19\right) = \frac59R_{\mathrm H}. \]

Therefore,

\[ \lambda_{\max} = \frac{9}{5(109\,678)}\ \mathrm{cm} = 1.64117\times10^{-5}\ \mathrm{cm}. \]

Using \(1\ \mathrm{cm}=10^8\ \text{Å}\),

\[ \boxed{\lambda_{\max}\approx1641\ \text{Å}}. \]

1.2 — Transitions to the ground state and its energy

Transitions terminating at \(n_f=1\) constitute the Lyman series:

\[ \boxed{ \frac{1}{\lambda} = R_{\mathrm H}\left(1-\frac{1}{n^2}\right), \qquad n=2,3,4,\ldots }. \]

At the series limit, \(n\rightarrow\infty\), the initial electron is at the zero-energy ionization threshold. Thus,

\[ 0-E_1=\frac{hc}{\lambda_{\mathrm{limit}}}=hcR_{\mathrm H}. \]

Consequently,

\[ E_1=-hcR_{\mathrm H} =-(1.23984198\times10^{-4})(109\,678)\ \mathrm{eV} =-13.5983\ \mathrm{eV}. \]

Hence,

\[ \boxed{E_1\approx-13.6\ \mathrm{eV}}. \]

The negative sign indicates that the electron is bound.

1.3 — Muonic hydrogen

With a fixed nucleus and an orbiting particle of mass \(m\) and charge \(-e\), the Bohr formulas give

\[ E_n=-\frac{m e^4}{8\varepsilon_0^2h^2n^2}, \qquad r_n=\frac{\varepsilon_0h^2n^2}{\pi m e^2}. \]

Thus the binding-energy magnitude is proportional to \(m\), while the radius is inversely proportional to \(m\). Substituting \(m_\mu=207m_e\),

\[ E_1^{(\mu)} = 207E_1^{(\mathrm H)} = 207(-13.5983)\ \mathrm{eV} \approx-2814.9\ \mathrm{eV}. \]

Therefore,

\[ \boxed{E_1^{(\mu)}\approx-2.81\times10^3\ \mathrm{eV}}. \]

The first-orbit radius is

\[ r_1^{(\mu)} = \frac{a_0}{207} = \frac{0.53}{207}\ \text{Å} = 2.5604\times10^{-3}\ \text{Å}. \]

To the precision of the supplied Bohr radius,

\[ \boxed{ r_1^{(\mu)}\approx2.6\times10^{-3}\ \text{Å} =2.6\times10^{-13}\ \mathrm m }. \]

1.4 — Probability within the spherical shell

The probability density is the squared magnitude of the wavefunction:

\[ |\Psi_{1s}(r)|^2 = \frac{1}{\pi a_0^3}e^{-2r/a_0}. \]

For a sufficiently thin shell,

\[ P\approx|\Psi_{1s}(r)|^2\,4\pi r^2\Delta r. \]

At \(r=a_0\), with \(\Delta r=0.001a_0\),

\[ \begin{aligned} P &\approx \frac{e^{-2}}{\pi a_0^3} \,4\pi a_0^2(0.001a_0)\\ &=0.004e^{-2}\\ &=5.4134\times10^{-4}. \end{aligned} \]

Thus,

\[ \boxed{P\approx5.41\times10^{-4}=0.0541\%}. \]

1.5 — Identification of the dissociation channels

The two balanced dissociation equations are

\[ \text{(i)}\qquad \ce{H2 -> H + H}, \]

\[ \text{(ii)}\qquad \ce{H2 -> H+ + H-}. \]

For channel (i), the energy of the two infinitely separated ground-state hydrogen atoms is

\[ E_\infty^{(i)} = 2E_1^{(\mathrm H)} \approx2(-13.6) =-27.2\ \mathrm{eV}. \]

This must be the lower asymptote. The upper asymptote, at \(-14.3\ \mathrm{eV}\), therefore corresponds to channel (ii).

Fill the upper box with (ii) and the lower box with (i).

For the ionic channel, the bare proton has zero electronic energy, so the upper asymptote also establishes

\[ E(\ce{H-})=-14.3\ \mathrm{eV}. \]

1.6 — Dissociation energies

The molecular potential-energy minimum is

\[ E_{\min}(\ce{H2})=-31.9\ \mathrm{eV}. \]

Each dissociation energy is the product asymptote minus this minimum.

For channel (i),

\[ \ce{H2 -> H + H}, \]

\[ D_e^{(i)} = -27.2-(-31.9) = \boxed{4.7\ \mathrm{eV}}. \]

For channel (ii),

\[ \ce{H2 -> H+ + H-}, \]

\[ D_e^{(ii)} = -14.3-(-31.9) = \boxed{17.6\ \mathrm{eV}}. \]

Enter these values in the corresponding channel (i) and channel (ii) answer boxes.

1.7 — Electron detachment from hydride

The balanced process is

\[ \ce{H- -> H + e-}. \]

Using zero energy for the detached free electron,

\[ \begin{aligned} \Delta E &=E(\ce{H})+E(\ce{e-})-E(\ce{H-})\\ &=-13.6+0-(-14.3)\\ &=\boxed{+0.7\ \mathrm{eV}}. \end{aligned} \]

The process requires \(0.7\ \mathrm{eV}\) per hydride ion.

1.8 — Effective nuclear charge of hydride

In the stipulated model, both electrons occupy the \(n=1\) level, and each has energy

\[ E_{\text{electron}}=-13.6Z_{\mathrm{eff}}^2\ \mathrm{eV}. \]

The total energy is therefore

\[ E(\ce{H-}) = 2(-13.6Z_{\mathrm{eff}}^2) = -27.2Z_{\mathrm{eff}}^2\ \mathrm{eV}. \]

Equating this to the energy obtained from the graph,

\[ -27.2Z_{\mathrm{eff}}^2=-14.3, \]

so

\[ \boxed{ Z_{\mathrm{eff}} = \sqrt{\frac{14.3}{27.2}} = 0.725 }. \]

Its value below \(1\) represents screening of the proton’s attraction by the other electron within this model.

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.