Sol. The given reaction is

P + Li 7 — → 2He 4 … (1)

The energy of this reaction is given as

Q = [(m p + m Li ) – 2m He ]c 2 … (2)

The α -particles produced possesses the total energy of (Q + T), where T is the kinetic energy of the proton. Since the α -particles fly symmetrically with respect to the original direction of motion of the proton, so the kinetic energy of both, α -particles is same. If the kinetic energy of one α -particle is T α , then applying energy conservation, we get,

2T α = Q + T … (3)

The reaction may be represented as shown in the figure.

So, applying momentum conservation we get,

2P α cos 2 ( θ /2) = P p

or, cos 2 ( θ /2) =

The momentum of the alpha particles and proton should be taken as the relativistic values to get more accurate results. So,

P α =

and, P p =

so, 4T α (T α + 2m α c 2 )cos 2 ( θ /2) = T(T + 2m p c 2 ) … (4)

So, from equations (3) and (4), we get

4{(Q + T)/2} {(Q + T)/2} + {2m α c 2 }× cos 2 ( θ /2) = T 2 + 2T p c 2 or, cos

2 ( θ /2) =

or, θ = 2cos –1

or, θ = 2cos –1 … (5)

Since the energy Q of the reaction and kinetic energy T are very small as compared to rest energies. So, above equation reduces to

or,

So, on putting the values, we get

And from the equation (3), we get

So, on putting the values, we get