Ncert Exercises & Solutions

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.

Hint: Use the law of conservation of momentum and the law of conservation of energy.

Step 1: Find the momentum of the neutron and proton.

Given binding energy, B = 2.2 MeV

From the energy conservation law,

E - B = K_{n} + K_{p} = \frac{p_{n}^{2}}{2 m} + \frac{p_{p}^{2}}{2 m} . . . . . (i)

From the conservation of momentum,

p_{n} + p_{p} = \frac{E}{c}

................(ii)

E = B, Eq . (i), p_{n}^{2} + p_{p}^{2} = 0

It can only happen if, p_n = p_p = 0

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

Step 2: Find the energy of proton and neutron.

Let E = B + X, where X<<<B

Put the value of p_n from Eq. (ii) in Eq. (i), we get,

X = \frac{{(\frac{E}{c} - p_{p})}^{2}}{2 m} + \frac{p_{p}^{2}}{2 m}

or 2 p_{p}^{2} - \frac{2 {Ep}_{p}}{c} + \frac{E^{2}}{c^{2}} - 2 mX = 0

Using the formula of quadratic equation, we get

p_{p} = \frac{\frac{2 E}{c} \pm \sqrt{\frac{4 E^{2}}{c^{2}} - 8 (\frac{E^{2}}{c^{2}} - 2 mX)}}{4}

For the real value p_p, the discriminant is positive

\frac{4 E^{2}}{c^{2}} = 8 (\frac{E^{2}}{c^{2}} - 2 mX)

16 mX = \frac{4 E^{2}}{c^{2}}

X = \frac{E^{2}}{4 {mc}^{2}} \approx \frac{B^{2}}{4 {mc}^{2}} [∴ X < < B \Rightarrow E ≅ B]

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