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## Homework Statement

I have a doubt in the following problem. I get an answer which is slightly different from the book answer.

1) A 6 MeV gamma ray is absorbed and dissociates a deuteron into a proton and a neutron. If the neutron makes an angle of 90 degrees with the direction of gamma ray, determine the kinetic energies of proton and neutron.

## Homework Equations

E=B.E + K(p) + K(n)

## The Attempt at a Solution

I solved it in the following way:

Let E be the energy of the gamma ray. Let K(p) and K(n) be the kinetic energies of proton and neutron respectively. Let m(p) and m(n) be the rest masses of proton and neutron respectively.

We assume that the gamma ray moves along positive X-direction, neutron moves along positive Y-direction and proton moves at an angle theta to the positive X axis.

Applying law of energy conservation,

E=B.E + K(p) + K(n)

Where B.E=Binding energy

K(p) + K(n) = 3.78 MeV --------(1)

Applying law of momentum conservation along X direction,

E/c = [2m(p)K(p)]^(1/2)cos(theta) ------------- (2)

Applying law of momentum conservation along Y direction,

0= [2m(n)K(n)]^(1/2) - [2m(p)K(p)]^(1/2)sin(theta)

=> [2m(n)K(n)]^(1/2) = [2m(p)K(p)]^(1/2)sin(theta) ---------- (3)

Squaring and adding equations (1) and (2) we get,

[E/c]^2 + 2m(n)K(n) = 2m(p)K(p)

On simplification I get the following equation,

K(p) – K(n) = 0.019 MeV ---------(4)

Solving equations (1) and (4),

K(p) = 1.89 MeV

K(n) = 1.87 MeV

But the answer is,

K(p) = 1.91 MeV

K(n) = 1.86 MeV

Please help.