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(a) Obtain the de Broglie wavelength of a neutron of kinetic energy $150 \mathrm{eV}$. As you have seen in question 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. $\left(m_{n}=1.675 \times 10^{-27} \mathrm{~kg}\right)$. (b) Obtain the de B roglie wavelength associated with thermal neutrons at room temperature (27 °C). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.
(a) Obtain the de Broglie wavelength of a neutron of kinetic energy $150 \mathrm{eV}$. As you have seen in question 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. $\left(m_{n}=1.675 \times 10^{-27} \mathrm{~kg}\right)$.
(b) Obtain the de B roglie wavelength associated with thermal neutrons at room temperature (27 °C). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.
CBSE | Class 12 | Dual Nature of Radiation and Matter | NCERT | Physics
(a) Obtain the de Broglie wavelength of a neutron of kinetic energy $150 \mathrm{eV}$. As you have seen in question 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. $\left(m_{n}=1.675 \times 10^{-27} \mathrm{~kg}\right)$.
(b) Obtain the de B roglie wavelength associated with thermal neutrons at room temperature (27 °C). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.

(a) Kinetic energy of the neutron is given as $=150 \mathrm{eV}$

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

$=2.4 \times 10^{-17} \mathrm{~J}$

Mass of the neutron, $m_{n}=1.675 \times 10^{-27} \mathrm{~kg}$

The kinetic energy of the neutron is given by the relation

$\mathrm{K} \cdot \mathrm{E}=(1 / 2) \mathrm{m}_{\mathrm{e}} \mathrm{V}^{2}$

$\mathrm{K} . \mathrm{E}=\mathrm{p}^{2} / 2 \mathrm{~m}_{\mathrm{e}}$

$\Rightarrow p=\sqrt{2 m_{e} K \cdot E}$

$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m_{c} K \cdot E}}$

The mass and wavelength are inversely related. With increasing mass, the wavelength decreases and vice versa.

$\lambda=\frac{\left(6.63 \times 10^{-34}\right)}{\sqrt{2.24 \times 10^{-17} \times 1.675 \times 10^{-27}}}$

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

The inter-atomic spacing of the crystal is around $1 \AA$, i.e., $10^{-10} \mathrm{~m}$, as indicated in question $11.31$. Atoms are separated by about 100 times their normal distance. As a result, diffraction investigations require a neutron with a kinetic energy of $150 \mathrm{eV}$.

(b) Room temperature $=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$

Average kinetic energy of the neutron, $E=(3 / 2) \mathrm{kT}$

here, $k=$ Boltzmann constant $=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{mol} / \mathrm{K}$

The wavelength of the neutron can be calculated as,

$\lambda=\frac{h}{\sqrt{2 m_{n} E}}$

$\lambda=\frac{h}{\sqrt{2 m_{n} \frac{3}{2} k T}}$

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

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

This wavelength is comparable to the crystal’s inter-atomic gap. As a result, before a rapid neutron beam can be employed for neutron diffraction investigations, it must be thermalized with the surroundings.

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