Sol. The given nuclear reaction may be represented as

C 12 + n — → Be 7 + α + Q … (1)

Here Q is the energy of the nuclear reaction and it is given by mass energy conservation. So,

m 1 + m 2 = m 3 + m 4 + Q

or, Q = m 1 + m 2 – m 3 – m 4 … (2)

If the masses of neutron and carbon are m 1 and m 2 respectively, then the energy Q in term of threshold energy is given as

T th = | Q |

It may be calculated by using above equation (2). It comes out to be negative.

So, Q = … (2)

Since the kinetic energy of incoming neutron is t, then total energy of the products of the reaction is Q + T. Now, if the kinetic energy of α -particle and the Be nucleus are T α and T Be , then,

Q + T = T α + T Be

or, T Be = Q + T – T α … (3)

This nuclear reaction may be represented as shown in the fig.

Now according to momentum conservation, we get

P n = P Be cos θ

and P α = P Be sin θ

So, on squaring adding above two equations, we get

+ = … (4)

Now, according to general equation, the momentum of these particles are given as

P n C =

or, P n =

Similarly,

P α =

and P Be =

Since the threshold kinetic energy T th and the kinetic energies of the particles in the reaction are very small as compared to their rest energies. So, the ratios T/2m 1 c 2 , T α /2m 3 c 2 and T Be /2m 4 c 2 may be neglected without appreciable loss in accuracy and the classical relation between momentum and kinetic energy may be used. So, we get,

= 2Tm 1

= 2T α m 3

= T Be m 4 … (5)

So, from equations (4) and (5), we get

2Tm 1 + 2T α m 3 = 2T Be m 4

or, Tm 1 + T α m 3 = T Be m 4 … (6)

So, from equations (3) and (6), we get

Tm 1 + T α m 3 = (Q + T – T α )m 4

or, T α m 3 + T α m 4 = Tm 4 + Qm 4 –Tm 1

since Q = m 2 T th /(m 1 + m 2 )

so, T α (m 3 + m 4 ) = T(m 4 – m 1 ) –

or, T α =

So, on putting the values, we get