Nuclear binding energy — UPSC Physics Optional CSE 2018 Paper II

Assuming that the neutron-proton interaction has a square well form V(r)=-V_0\quad\text{for }r\leq b =0\quad\text{for }r>b, the ground state wave function of deuteron nucleus is given as \psi(r)=A\sin kr\quad\text{for }r\leq b =Ce^{-\gamma r}\quad\text{for }r>b where k=\sqrt{\frac{M}{\hbar^2}(V_0+W)} and \gamma=\sqrt{\frac{MW}{\hbar^2}}. Here M is the nucleon mass, W is the binding energy of deuteron and A and C are constants.

(i) Show that for a just bound state of deuteron V_0b^2=\frac{\pi^2\hbar^2}{4M}.

(ii) Explain why deuteron is a loosely bound extended structure.