Question

(a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exercise 11.31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain.($m_n=1.675\times {10}^{27}$kg)
(b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (${27}^{o}C$). Hence, explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.

Exercise 11.31:

Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wave length of the probe equal to 1 A∘A∘ which is of the order of inter-atomic spacing in the lattice) ($m_e=9.11\times {10}^{-31}kg$)

Answer 1

(a)The De Broglie wavelength of the neutron is $=2.327\times {10}^{-12}m$.

Kinetic energy of the neutron, $K=150eV$

$=150\times 1.6\times {10}^{-19}$

$=2.4\times {10}^{-19}J$

Mass of a neutron, $m_e=1.675\times {10}^{-27}kg$

The relation that gives the kinetic energy of neutron is:.

$K=\frac{1}{2}{m}_e{v}^2$

$m_{e}v=\sqrt{K{m}_e}$

De-Broglie wavelength of the neutron is given as;

$\lambda =\frac{h}{m_{e}v}$

$\lambda =\frac{h}{\sqrt{2K{m}_e}}$

The wavelength of the neutron is inversely proportional to the square root of its mass. So, there will be a decrease in wavelength with increase in mass and vice versa.

$\therefore \lambda =\frac{6.6\times {10}^{-34}}{\sqrt{2\times 2.4\times {10}^{-17}\times 1.675\times {10}^{-27}}}$

$=2.327\times {10}^{-12}m$

So, the neutron beam of the same energy $\left(150eV\right)$ will not be suitable.

(b)

Room temperature, $T={27}^{0}C=27+273=300K$

The relation for average kinetic energy of the neutron is:

$E=\frac{3}{2}kT$

The value of $k=$ Boltzmann constant $=1.38\times {10}^{-23}Jmo{l}^{-1}{K}^{-1}$

The wavelength of the neutron is given as:

$\lambda =\frac{h}{\sqrt{2m_{e}E}}=\frac{h}{\sqrt{3m_{e}kT}}$

$=\frac{6.6\times {10}^{-34}}{\sqrt{3\times 1.675\times {10}^{-27}\times 1.38\times {10}^{-23}\times 300}}$

$=1.447\times {10}^{-10}m$

The wavelength is comparable to the inter atomic spacing of a crystal, Hence, the high energy neutron beam should first be thermallsed, before using it for diffraction.