Question

Deuteron is a bound state of a neutron and a proton with a binding energy $B = 2.2{\rm{ MeV}}$. A $\gamma$-ray of energy E is aimed at a deuteron nucleus to try to break it into a (neutron + proton) such that the n and p move in the direction of the incident $\gamma$-ray. If $E = B$, show that this cannot happen. Hence, calculate how much bigger than B must be E be for such a process to happen?

Answer 1

We will use the concept of law of momentum conservation and energy conservation for the given process when the nucleus of the deuteron breaks into neutrons and protons. We will assume the value of B greater than E by some factor and the value of that factor will be determined.

Complete step by step answer:
From the concept of law of conservation of energy for neutron and proton, we can say that the difference of energy of ray and binding energy is equal to the summation of the kinetic energy of neutron and proton.
$E - B = {K_n} + {K_p}$……(1)
Here, ${K_n}$ and ${K_p}$ are kinetic energy of neutrons and protons respectively.
We know that the relation between kinetic energy and momentum of the neutron is given as:
${K_n} = \dfrac{{P_n^2}}{{2{m_n}}}$
Here, ${P_n}$ is momentum and ${m_n}$is the mass of the neutron.
Also, the relation between kinetic energy and momentum of the proton can be written as:
${K_p} = \dfrac{{P_p^2}}{{2{m_p}}}$
Here, ${P_p}$ is momentum and ${m_p}$is the mass of the proton.
Substitute $\dfrac{{P_n^2}}{{2{m_n}}}$ for ${K_n}$ and $\dfrac{{P_p^2}}{{2{m_p}}}$ for ${K_p}$ in equation (1).
$E - B = \dfrac{{P_n^2}}{{2{m_n}}} + \dfrac{{P_p^2}}{{2{m_p}}}$……(2)
We can assume the mass of the neutron and proton are equal.

{m_n} = {m_p}\\\ = m \end{array}$$ Substitute m for $${m_n}$$ and $${m_p}$$ in equation (2). $$E - B = \dfrac{{P_n^2}}{{2m}} + \dfrac{{P_p^2}}{{2m}}$$……(3) From the concept of law of conservation of momentum for neutrons and protons, we can say that the summation of the momentum of neutrons and protons is equal to the initial momentum of the ray. $$\begin{array}{l} {P_n} + {P_p} = \dfrac{E}{c}\\\ {P_n} = \dfrac{E}{c} - {P_p} \end{array}$$ Here, $$\dfrac{E}{c}$$ is the initial momentum. On substituting E for B in equation (3), we get: $$\begin{array}{l} E - E = \dfrac{{P_n^2}}{{2m}} + \dfrac{{P_p^2}}{{2m}}\\\ 0 = \dfrac{{P_n^2}}{{2m}} + \dfrac{{P_p^2}}{{2m}} \end{array}$$ Based on the above expression, we can conclude that the momentum of the neutron and proton is zero, which means this process cannot happen. Let us assume that the value of E is greater than E by a constant $$\lambda $$. $$E = B + \lambda $$ On substituting $$B + \lambda $$ for E for in equation (3), we get: $$\begin{array}{l} E - \left( {B + \lambda } \right) = \dfrac{{P_n^2}}{{2m}} + \dfrac{{P_p^2}}{{2m}}\\\ \lambda = \dfrac{{P_n^2}}{{2m}} + \dfrac{{P_p^2}}{{2m}} \end{array}$$ Substitute $$\dfrac{E}{c} - {P_p}$$ for $${P_n}$$ in the above expression. $$\begin{array}{l} \lambda = \dfrac{{{{\left( {\dfrac{E}{c} - {P_p}} \right)}^2}}}{{2m}} + \dfrac{{P_p^2}}{{2m}}\\\ = \dfrac{1}{{2m}}\left[ {P_p^2 + {{\left( {{P_p} - \dfrac{E}{c}} \right)}^2}} \right]\\\ 0 = 2P_p^2 - 2\dfrac{E}{c}{P_p} + \left( {\dfrac{E}{{{c^2}}} - 2m\lambda } \right) \end{array}$$ On solving the above expression, we can write: $${P_p} = \dfrac{{2\dfrac{E}{c} \pm \sqrt {4\dfrac{{{E^2}}}{{{c^2}}} - 8\left( {\dfrac{{{E^2}}}{{{c^2}}} - 2m\lambda } \right)} }}{4}$$ For the real value of momentum of the proton, the discriminant term in the above expression should be greater than or equal to zero. $$\begin{array}{l} \sqrt {4\dfrac{{{E^2}}}{{{c^2}}} - 8\left( {\dfrac{{{E^2}}}{{{c^2}}} - 2m\lambda } \right)} \ge 0\\\ 16m\lambda = 4\dfrac{{{E^2}}}{{{c^2}}}\\\ \lambda = \dfrac{{{E^2}}}{{4m{c^2}}} \end{array}$$ Considering the value of E approximately equal to binding energy B, we can write the above expression as below: $$\lambda = \dfrac{{{B^2}}}{{4m{c^2}}}$$ Therefore, for the given process to occur the value of energy E must be greater than binding energy B by a factor $$\lambda $$ which is equal to $$\dfrac{{{B^2}}}{{4m{c^2}}}$$. **Note:** In this problem, it would be an added advantage if we remember the Sridharacharya formula to find the roots of a quadratic equation. Also, students need to take extra care while rearranging the equations.