Question

A 2 Mev neutron is scattered elastically by 12C through an angle of 45 degrees. (a) What is the scattered neutron’s kinetic energy? (b) What is the recoil energy of the 12C nucleus? (c) At what angle does the recoiling nucleus appear?

Answer 1

Answer:

(a) 1.716 MeV

(b) 0.284 MeV

(c) $45^{\circ}$

Solution:

As per the question:

Energy of the scattered neutron, E = 2 MeV = $2\times 10^{6}\ eV$

Atomic no. of $^{12}\textrm{C},\ A = 12$

Now,

To calculate the Kinetic Energy of the scattered neutron, $E_{av} = \frac{1}{2}(1 - \alpha)E$

where

$\alpha = (\frac{A - 1}{A + 1})^{2}$

Therefore,

$E_{av} = \frac{1}{2}(1 - (\frac{A - 1}{A + 1})^{2})E$

$E_{av} = \frac{1}{2}(1 - (\frac{12 - 1}{12 + 1})^{2})\times 2\times 10^{6} = 0.284\ MeV$

(a) The energy of the scattered neutron is: 2 MeV - 0.284 MeV = 1.716 MeV

(b) The recoil energy of the $^{12}\textrm{C}$ nucleus is 0.284 MeV

(c) The angle of the recoil of the nucleus is $45^{\circ}$