The fundamental problem in harnessing nuclear fusion is getting the two particles (say, two deuterons) close enough together for the attractive (but short-range) nuclear force to overcome the Coulomb repulsion. The “bulldozer” method is to heat the particles up to fantastic temperatures and allow the random collisions to bring them together. A more exotic proposal is muon catalysis, in which we construct a “hydrogen molecule ion,” only with deuterons in place of protons, and a muon in place of the electron. Predict the equilibrium separation distance between the deuterons in such a structure, and explain why muons are superior to electrons for this purpose.

The equilibrium separation distance between the deuterons but$m_e\to m_r$the reduced mass of themuon:

role="math" localid="1658383790270" $$\begin{gathered} m_r=\frac{m_{\mu }{m}_d}{m_{\mu }+m_d} \\ =\frac{m_{\mu }2m_p}{m_{\mu }+2m_p} \\ =\frac{m_{\mu }}{1+m_{\mu }/2m_p}. \\ {m}_{\mu }=207m_e, \end{gathered}$$

so

$$\begin{gathered} 1+\frac{m_{\mu }}{m_p}=1+\frac{207}{2}\frac{\left(9.11\times {10}^{-31}\right)}{\left(1.67\times {10}^{-27}\right)} \\ =1.056;m_r \\ =\frac{207m_e}{1.056}=196m_e \end{gathered}$$

This shrinks the muonic \Bohr radius” down by a factor of nearly 200. but nowis the ground state energy of the muonic atom. The potential energy associated with the deuteron-deuteron repulsion is the same as, and since the productis independent of mass, it doesn’t matter whether we write it using the electron values or the muon values. the entire molecule shrinks by that same factor of 196. The equilibrium separation for the electron case was 2.493 a (Problem 7.10),Therefore, The general equation of

$$\begin{gathered} R=\frac{2.493}{196}\left(0.529\times {10}^{-10}m\right) \\ =6.73\times {10}^{-10}m \end{gathered}$$

Thus the equilibrium separation distance between deuterons is $R=6.73\times {10}^{-10}m.$