IChO 2010 · Theory problem 5

Photoelectron spectroscopy

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When an atom \(\ce{X}\) absorbs radiation with a photon energy greater than the ionization energy of the atom, the atom is ionized to generate an ion \(\ce{X+}\) and the electron (called a photoelectron) is ejected at the same time. In this event, the energy is conserved as shown in Figure 1, that is,

Photon energy (\(h\nu\)) = ionization energy (IE) of \(\ce{X}\) + kinetic energy of photoelectron.

When a molecule, for example, \(\ce{H2}\), absorbs short-wavelength light, the photoelectron is ejected and an \(\ce{H2+}\) ion with a variety of vibrational states is produced. A photoelectron spectrum is a plot of the number of photoelectrons as a function of the kinetic energy of the photoelectrons. Figure 2 shows a typical photoelectron spectrum when \(\ce{H2}\) in the lowest vibrational level is irradiated by monochromatic light of 21.2 eV. No photoelectrons are detected above 6.0 eV. (eV is a unit of energy and 1.0 eV is equal to \(1.6\cdot10^{-19}\) J.)

Figure 1

Figure 1. Schematic diagram of photoelectron spectroscopy.

Figure 2

Figure 2. Photoelectron spectrum of \(\ce{H2}\). The energy of the incident light is 21.2 eV.

5.1 a) Determine the energy difference \(\Delta E_{\mathrm{A1}}\) (eV) between \(\ce{H2}\) (\(v=0\)) and \(\ce{H2+}\) (\(v_{\mathrm{ion}}=0\)) to the first decimal place. \(v\) and \(v_{\mathrm{ion}}\) denote the vibrational quantum numbers of \(\ce{H2}\) and \(\ce{H2+}\), respectively.

b) Determine the energy difference \(\Delta E_{\mathrm{A2}}\) (eV) between \(\ce{H2+}\) (\(v_{\mathrm{ion}}=0\)) and \(\ce{H2+}\) (\(v_{\mathrm{ion}}=3\)) to the first decimal place.

The electronic energy levels \(E_n^{\mathrm{H}}\) of a hydrogen atom are given by the equation

\[ E_n^{\mathrm{H}}=-\frac{Ry}{n^2} \qquad (n=1,2,3\cdots) \]

Here \(n\) is a principal quantum number, and \(Ry\) is a constant with dimensions of energy. The energy from \(n=1\) to \(n=2\) of the hydrogen atom is 10.2 eV.

5.2 Calculate the ionization energy \(E_{\mathrm{B}}\) (eV) of the hydrogen atom to the first decimal place.

The energy threshold for the generation of two electronically excited hydrogen atoms \(\ce{H^{*}}\) (\(n=2\)) from \(\ce{H2}\) (\(v=0\)) has been derived to be 24.9 eV by an experiment.

5.3 Determine the bond energy \(E_{\mathrm{C}}\) (eV) of \(\ce{H2}\) to the first decimal place.

5.4 Considering an energy cycle, determine the bond energy \(E_{\mathrm{D}}\) (eV) of \(\ce{H2+}\) to the first decimal place. If you were unable to determine the values for \(E_{\mathrm{B}}\) and \(E_{\mathrm{C}}\), then use 15.0 eV and 5.0 eV for \(E_{\mathrm{B}}\) and \(E_{\mathrm{C}}\), respectively.

5.5 Calculate the threshold energy \(E_{\mathrm{E}}\) (eV) of the following dissociative ionization reaction to the first decimal place:

Figure 3

If you were unable to determine the values for \(E_{\mathrm{B}}\) and \(E_{\mathrm{C}}\), then use 15.0 eV and 5.0 eV for \(E_{\mathrm{B}}\) and \(E_{\mathrm{C}}\), respectively.

When \(\ce{H2}\) absorbs monochromatic light of 21.2 eV, the following dissociation process occurs at the same time.

Figure 4

Two hydrogen atoms move in opposite directions with the same speed.

5.6 Calculate the speed \(u\) (\(\mathrm{m\,s^{-1}}\)) of the hydrogen atoms generated in the above reaction. \(\ce{H2}\) is assumed to be at rest. If you were unable to determine the value for \(E_{\mathrm{C}}\), then use 5.0 eV for \(E_{\mathrm{C}}\).

Official English text of the problem.

Answers

  • 5.1 a) \(\boxed{\Delta E_{\mathrm{A1}}=15.4\ \mathrm{eV}}\).
  • 5.1 b) \(\boxed{\Delta E_{\mathrm{A2}}=0.8\ \mathrm{eV}}\).
  • 5.2 \(\boxed{E_{\mathrm{B}}=13.6\ \mathrm{eV}}\).
  • 5.3 \(\boxed{E_{\mathrm{C}}=4.5\ \mathrm{eV}}\).
  • 5.4 \(\boxed{E_{\mathrm{D}}=2.7\ \mathrm{eV}}\).
  • 5.5 \(\boxed{E_{\mathrm{E}}=28.3\ \mathrm{eV}}\).
  • 5.6 \(\boxed{u=4.0\times10^4\ \mathrm{m\,s^{-1}}}\) for each atom, using \(m_{\mathrm H}\approx1.67\times10^{-27}\ \mathrm{kg}\).

Solution

The species used below are molecular hydrogen (dihydrogen), \(\ce{H2}\), [H][H]; the hydrogen molecular ion (dihydrogen radical cation), \(\ce{H2+}\), [H][H+]; atomic hydrogen, \(\ce{H}\), [H]; and the hydrogen ion (proton), \(\ce{H+}\), [H+]. For \(\ce{H2+}\), the SMILES is a connectivity and charge shorthand: it does not faithfully represent the delocalized one-electron bond, whose bond order is \(1/2\). These SMILES also do not specify vibrational states or atomic electronic excitation.

5.1 a)

Photoionization produces an ion and a photoelectron:

\[ \ce{H2 + h\nu -> H2+ + e-}. \]

The initial molecule has \(v=0\). If the ion is produced in vibrational level \(j\), conservation of energy gives

\[ h\nu=\Delta E_{\mathrm{A1}}+\epsilon_j+K_j, \]

where \(\epsilon_j\) is the ion’s vibrational energy above \(v_{\mathrm{ion}}=0\), and \(K_j\) is the photoelectron kinetic energy.

The greatest photoelectron kinetic energy therefore corresponds to the lowest final ionic energy, \(v_{\mathrm{ion}}=0\). The kinetic-energy axis decreases from left to right, so this is the leftmost peak, centered at approximately \(5.8\ \mathrm{eV}\).

Thus,

\[ \boxed{\Delta E_{\mathrm{A1}} =21.2-5.8 =15.4\ \mathrm{eV}}. \]

5.1 b)

Successive peaks toward lower kinetic energy correspond to successively higher ionic vibrational levels. Counting the first peak as \(v_{\mathrm{ion}}=0\), the fourth peak represents \(v_{\mathrm{ion}}=3\). Its kinetic energy is approximately \(5.0\ \mathrm{eV}\).

Because the photon energy and initial molecular state are identical, the increase in ionic energy equals the decrease in photoelectron kinetic energy:

\[ \begin{aligned} \Delta E_{\mathrm{A2}} &=E_{\mathrm{ion}}(3)-E_{\mathrm{ion}}(0)\\ &=K_0-K_3\\ &=5.8-5.0. \end{aligned} \]

Therefore,

\[ \boxed{\Delta E_{\mathrm{A2}}=0.8\ \mathrm{eV}}. \]

5.2

For the given electronic energy levels,

\[ \begin{aligned} E_2^{\mathrm H}-E_1^{\mathrm H} &=-\frac{Ry}{4}-(-Ry)\\ &=\frac34Ry =10.2\ \mathrm{eV}. \end{aligned} \]

Hence,

\[ Ry=\frac43(10.2)=13.6\ \mathrm{eV}. \]

Ionization of ground-state atomic hydrogen is

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

The free-electron limit is \(n\to\infty\), for which \(E_\infty^{\mathrm H}=0\). Therefore,

\[ \boxed{E_{\mathrm B} =E_\infty^{\mathrm H}-E_1^{\mathrm H} =Ry =13.6\ \mathrm{eV}}. \]

5.3

The experimentally specified process is

\[ \ce{H2 -> 2H^{*}}, \qquad n=2\text{ for both product atoms}. \]

It can be resolved into dissociation to ground-state atoms,

\[ \ce{H2 -> 2H}, \qquad \text{energy required }E_{\mathrm C}, \]

followed by excitation of both atoms:

\[ \ce{H -> H^{*}}, \qquad n=1\to2, \qquad 10.2\ \mathrm{eV}\text{ per atom}. \]

Consequently,

\[ 24.9\ \mathrm{eV}=E_{\mathrm C}+2(10.2\ \mathrm{eV}), \]

and

\[ \boxed{E_{\mathrm C}=24.9-20.4=4.5\ \mathrm{eV}}. \]

This is the dissociation energy from the molecular \(v=0\) level.

5.4

The required ionic dissociation is

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

with the ion initially in \(v_{\mathrm{ion}}=0\) and the hydrogen atom produced in its ground electronic state.

Consider two routes from ground-state \(\ce{H2}\) to the same products, \(\ce{H + H+ + e-}\).

The first route dissociates the molecule and then ionizes one atom:

\[ \begin{aligned} \ce{H2 -> 2H} &\qquad E_{\mathrm C},\\ \ce{H -> H+ + e-} &\qquad E_{\mathrm B}. \end{aligned} \]

Its total energy requirement is \(E_{\mathrm C}+E_{\mathrm B}\).

The second route ionizes the molecule and then dissociates the molecular ion:

\[ \begin{aligned} \ce{H2 -> H2+ + e-} &\qquad \Delta E_{\mathrm{A1}},\\ \ce{H2+ -> H + H+} &\qquad E_{\mathrm D}. \end{aligned} \]

Conservation of energy around the cycle requires

\[ E_{\mathrm C}+E_{\mathrm B} =\Delta E_{\mathrm{A1}}+E_{\mathrm D}. \]

Thus,

\[ \boxed{E_{\mathrm D} =4.5+13.6-15.4 =2.7\ \mathrm{eV}}. \]

5.5

The dissociative ionization reaction is

\[ \ce{H2 -> H^{*} + H+ + e-}, \qquad \ce{H^{*}}\text{ has }n=2. \]

Starting from \(\ce{H2}\) in \(v=0\), the required energy is the sum of:

  1. Dissociation into two ground-state hydrogen atoms: \(E_{\mathrm C}=4.5\ \mathrm{eV}\).
  2. Excitation of one atom to \(n=2\): \(10.2\ \mathrm{eV}\).
  3. Ionization of the other atom: \(E_{\mathrm B}=13.6\ \mathrm{eV}\).

At threshold, no excess energy is available for product kinetic energy. Therefore,

\[ \boxed{E_{\mathrm E} =E_{\mathrm C}+10.2+E_{\mathrm B} =4.5+10.2+13.6 =28.3\ \mathrm{eV}}. \]

5.6

The photodissociation is

\[ \ce{H2 + h\nu -> 2H}, \qquad h\nu=21.2\ \mathrm{eV}, \qquad n=1\text{ for both atoms}. \]

After supplying the bond dissociation energy, the remaining photon energy becomes the total kinetic energy:

\[ K_{\mathrm{total}} =h\nu-E_{\mathrm C} =21.2-4.5 =16.7\ \mathrm{eV}. \]

Both atoms have the same mass and speed, so

\[ K_{\mathrm{total}} =2\left(\frac12m_{\mathrm H}u^2\right) =m_{\mathrm H}u^2. \]

Using the stated conversion \(1.0\ \mathrm{eV}=1.6\times10^{-19}\ \mathrm J\), together with the standard hydrogen-atom mass \(m_{\mathrm H}\approx1.67\times10^{-27}\ \mathrm{kg}\),

\[ \begin{aligned} u &=\sqrt{\frac{(16.7\ \mathrm{eV}) (1.6\times10^{-19}\ \mathrm{J\,eV^{-1}})} {1.67\times10^{-27}\ \mathrm{kg}}}\\[2mm] &=\boxed{4.0\times10^4\ \mathrm{m\,s^{-1}}}. \end{aligned} \]

Each atom has this speed, with the two velocities pointing in opposite directions.

The solution was checked against the official solution.