Question

A meutron collides, head-on with a deuterium at rest. What fraction of the neutron’s energy would be transferred to the deuterium?

• A. 89%
• B. 11%
• C. 79%
• D. 21%

Answer 1

Answer: (A)

89%

Answer 2

According to law of conservation of linear momentum, we get

$m_{n}v_{ni} + m_{d} \times 0 = m_{n}v_{nf} + m_{d}v_{df}$

Where $v_{ni}$is the initial velocity of the neutron before collision and $v_{nf}$and $v_{df}$are the velocities of the neutron and deuterium after collision

$m_{n}v_{ni} = m_{n}v_{nf} + m_{d}v_{df}$ (i)

According to conservation of kinetic energy, we get

$\frac{1}{2}m_{n}v_{ni}^{2} = \frac{1}{2}m_{n}v_{nf}^{2} + \frac{1}{2}m_{d}v_{df}^{2}$ ……(ii)

From eqs. (i) and (ii), it follows that

$m_{n}v_{ni}(v_{df} - v_{ni}) = m_{n}v_{nf}(v_{df} - v_{nf})$

$v_{df}(v_{ni} - v_{nf}) = v_{ni}^{2} - v_{nf}^{2}$

$v_{df} = v_{ni} + v_{nf}$

Substituting this in Eq. (i), we get

$v_{nf} = \frac{(m_{n} - m_{d})}{m_{n} + m_{d}}v_{ni}$ ……(iii)

$andv_{df} = \frac{2m_{n}v_{ni}}{m_{n} + m_{d}}$ ……..(iv)

The initial kinetic energy of the neutron is

$k_{ni} = \frac{1}{2}m_{n}v_{ni}^{2}$

Final kinetic energy of the deuterium is

$K_{df} = \frac{1}{2}m_{d}v_{df}^{2} = \frac{1}{2}m_{d}\left( \frac{2m_{n}v_{ni}}{m_{n} + m_{d}} \right)$ (Using (iv))

Fraction of neutron’s energy transferred to deuterium is

$f = \frac{K_{df}}{K_{ni}} = \frac{4m_{n}m_{d}}{(m_{n} + m_{d})^{2}}$

For deuterium, $m_{d} = 2m_{n}$

$\therefore f = \frac{4(m_{n})(2m_{n})}{(m_{n} + 2m_{n})^{2}} = \frac{8}{9} = 89\%$