. Show that a photon cannot spontaneously disintegrate into an electron- positron pair. [Hint: four-momentum conservation.] But in the presence of a stationary nucleus (acting as a kind of catalyst) it can. If the rest mass of the nucleus is \(N\), and that of the electron (and positron) is \(m\), what is the threshold frequency of the photon? Verify that for large \(N\) the efficiency is \(\sim 100\) per cent (cf. the preceding problem), so that the nucleus then comes close to being a pure catalyst.

We need to check if energy and momentum conservation laws are upheld for spontaneous disintegration of a photon into an electron-positron pair. The energy conservation equation is: $$ E_{\gamma} = E_{e^-} + E_{e^+} $$ The momentum conservation equation is: $$ \vec{p}_{\gamma} = \vec{p}_{e^-} + \vec{p}_{e^+} $$ where \(E_{\gamma}\) and \(\vec{p}_{\gamma}\) are the energy and momentum of the photon, and \(E_{e^-}\), \(E_{e^+}\), \(\vec{p}_{e^-}\) and \(\vec{p}_{e^+}\) are those of the electron and positron. For the photon, we have \(E_{\gamma}=p_{\gamma}c\) and for each of electron and positron \(E_{\text{particle}}^2 = (m_{\text{particle}}c^2)^2+(\vec{p}_{\text{particle}}c)^2\) Since the photon has no mass, its four-momentum conservation equation is not upheld in this case, and the photon cannot spontaneously disintegrate into an electron-positron pair.

In the presence of a stationary nucleus, we have a photon, a nucleus, an electron, and a positron. We can write the energy and momentum conservation equations as: $$ E_{\gamma} + M_Nc^2 = E_{e^-} + E_{e^+} + M_N^{'}c^2 $$ and $$ \vec{p}_{\gamma} = \vec{p}_{e^-} + \vec{p}_{e^+} + \vec{p}_N^{'} $$ Where \(M_N\) and \(M_N^{'}\) are the rest mass of the nucleus before and after the interaction, respectively. To satisfy momentum conservation, the final sum of the nucleus and electron-positron pair momenta should be equal to the initial photon momentum. As a result, this process can occur.

To find the threshold frequency of the photon, we will analyze the energy conservation equation, where the nucleus's kinetic energy (and thus momentum change) is minimized: $$ E_{\gamma} = E_{e^-} + E_{e^+} + (M_N^{'}-M_N)c^2 $$ In the threshold condition, we have \(E_{\text{particle}}=m_{\text{particle}}c^2\) for electron and positron $$ E_{\gamma} = 2mc^2 + (M_N^{'}-M_N)c^2 $$ The photon's energy is \(E_{\gamma} = h\nu\), with \(\nu\) being the frequency and \(h\) being the Planck constant. The threshold frequency is: $$ \nu = \frac{2mc^2 + (M_N^{'}-M_N)c^2}{h} $$

The efficiency of the process can be defined as: $$ \eta = \frac{2mc^2}{E_{\gamma}} $$ In the limit of large N, we have \(M_N^{'} \approx M_N\) $$ \eta = \frac{2mc^2}{2mc^2 + (M_N^{'}-M_N)c^2} \approx 1 $$ Therefore, for large N, the efficiency is close to 100%, and the nucleus comes close to being a pure catalyst.