IPhO 1974 · Theory problem 1

Inelastic collision of two hydrogen atoms (Bohr model, Doppler shift)

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A hydrogen atom in the ground state, moving with velocity \(v\), collides with another hydrogen atom in the ground state at rest. Using the Bohr model find the smallest velocity \(v_0\) of the atom below which the collision must be elastic.

At velocity \(v_0\) the collision may be inelastic and the colliding atoms may emit electromagnetic radiation. Estimate the difference of frequencies of the radiation emitted in the direction of the initial velocity of the hydrogen atom and in the opposite direction as a fraction (expressed in percents) of their arithmetic mean value.

Data:

\[ E_i=\frac{me^4}{2\hbar^2}=13.6\ \mathrm{eV} =2.18\cdot18^{-18}\ \mathrm{J}; \quad\text{(ionization energy of hydrogen atom)} \]

\[ m_H=1.67\cdot10^{-27}\ \mathrm{kg}; \quad\text{(mass of hydrogen atom)} \]

(\(m\) - mass of electron; \(e\) - electric charge of electron; \(\hbar\) - Planck constant; numerical values of these quantities are not necessary.)

Official English text of the problem.

Answers

The original has two unnumbered questions and gives no point values.

  • First question — threshold speed: \[ \boxed{v_0=\sqrt{\frac{3E_i}{m_H}} \simeq6.26\times10^4\ \mathrm{m\,s^{-1}}.} \]

  • Second question — frequency difference: With \(\nu_+\) denoting forward radiation and \(\nu_-\) backward radiation, \[ \boxed{\frac{\nu_+-\nu_-}{(\nu_++\nu_-)/2} \simeq\frac{v_0}{c} =2.09\times10^{-4}=0.0209\%.} \] The forward radiation has the higher frequency.

Solution

First question: threshold for an inelastic collision

In the Bohr model, the hydrogen energy levels are \[ E_n=-\frac{E_i}{n^2}. \] The smallest possible atomic excitation takes one atom from \(n=1\) to \(n=2\), requiring \[ \Delta E=E_2-E_1 =E_i\left(1-\frac14\right) =\frac34E_i =10.2\ \mathrm{eV}. \]

Both energy and momentum must be conserved. In particular, the entire initial kinetic energy cannot become excitation energy, because the atoms must retain the initial total momentum.

Let \(\mathbf v_1\) and \(\mathbf v_2\) be their velocities after the collision. Neglecting the tiny change in atomic mass upon excitation, momentum conservation gives \[ m_H\mathbf v=m_H\mathbf v_1+m_H\mathbf v_2. \] Their final translational kinetic energy can therefore be written as \[ K_f=\frac12m_H(v_1^2+v_2^2) =\frac14m_Hv^2 +\frac14m_H|\mathbf v_1-\mathbf v_2|^2. \] Its minimum is \[ K_{f,\min}=\frac14m_Hv^2, \] attained when both atoms move together with velocity \(\mathbf v/2\).

Thus the maximum energy available for excitation is \[ K_i-K_{f,\min} =\frac12m_Hv^2-\frac14m_Hv^2 =\frac14m_Hv^2. \] The excitation threshold consequently satisfies \[ \frac14m_Hv_0^2=\Delta E=\frac34E_i, \] so \[ v_0=\sqrt{\frac{3E_i}{m_H}}. \]

Using \(E_i=2.18\times10^{-18}\ \mathrm J\), consistent with the given \(13.6\ \mathrm{eV}\), \[ v_0= \sqrt{\frac{3(2.18\times10^{-18})}{1.67\times10^{-27}}} \simeq\boxed{6.26\times10^4\ \mathrm{m\,s^{-1}}}. \]

Below this speed, no atomic excitation is energetically possible, so the collision is elastic within the intended Bohr-level model.

Second question: Doppler frequency difference

At threshold, the atoms have no relative kinetic energy after excitation. The excited atom therefore moves in the initial direction with speed \[ u=\frac{v_0}{2}\simeq3.13\times10^4\ \mathrm{m\,s^{-1}}. \]

When it returns to the ground state, its rest-frame emission frequency is \[ \nu_*=\frac{\Delta E}{h}, \qquad h=2\pi\hbar, \] neglecting photon recoil.

Since \(u/c\simeq1.04\times10^{-4}\ll1\), the first-order Doppler formulas suffice: \[ \nu_+\simeq\nu_*\left(1+\frac{u}{c}\right), \qquad \nu_-\simeq\nu_*\left(1-\frac{u}{c}\right). \] Hence \[ \nu_+-\nu_-\simeq2\nu_*\frac{u}{c}, \qquad \frac{\nu_++\nu_-}{2}\simeq\nu_*. \] Dividing eliminates the rest-frame frequency: \[ \frac{\nu_+-\nu_-}{(\nu_++\nu_-)/2} \simeq\frac{2u}{c} =\frac{v_0}{c}. \]

With \(c\simeq3.00\times10^8\ \mathrm{m\,s^{-1}}\), the requested percentage is \[ \frac{6.26\times10^4}{3.00\times10^8}\times100\% \simeq\boxed{0.0209\%}. \]

The solution was checked against the official solution.