Question

Deuteron is a bound state of a neutron and a proton with a binding energy B = 2.2 MeV. A γ-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 γ-ray. If E = B, show that this cannot happen. Hence calculate how much bigger than B must E be for such a process to happen.

Answer 1

Given the binding energy of a deuteron, B = 2.2 MeV

Let kinetic energy and momentum of neutron and proton be K_n, K_P and p_n, p_p respectively.

From the conservation of energy,

`E - B = K_n + K_p = p_n^2/(2m) + p_p^2/(2m)` ......(i)

Now applying conservation of momentum,

`p_n + p_p = E/C`  ......(ii)

As E = B, equation (i) `p_n^2 + p_p^2` = 0

It only happens if p_n = p_p.

So, the equation. (ii) cannot be satisfied and the process cannot take place.

Let us take E = B + x, where x << B for the process to take place.

Putting the value of p, from equation (ii) in equation (i), we get

or `2p_p^2 - ((2E)/c)p_p + (E^2/c^2 - 2mx)` = 0

Solving the quadratic equation, we get

`p_p = (2E)/c + sqrt((4E^2)/c^2 - 8(E^2/c^2 - 2mx))/4`

For the real value p_p the discriminant is positive

`(4E^2)/c^2 = 8(E^2/c^2 - 2mx)`

`16mx = (4E^2)/c^2`

⇒ `x = E^2/(4mc^2)`

But x << B, hence E ≅ B

⇒ `x ≈ B^2/(4mc^2)`