physics - 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 * 10^{-27} kg). (b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27 ºC). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for n

Swetha (last edited 3 months ago) (Student, UG1 Grade)

Given that

Kinetic energy of neutron = 150 eV

= 150 * 1.6* 10^{-19} joules

Mass of neutron m_n = 1.675* 10^{-27} kg

Step 1: Determining the de Broglie wavelength of neutron

Recall the formula of momentum and kinetic energy

Momentum p = mv

Kinetic energy K.E = \frac{1}{2} mv^2

Where, p - momentum, m - mass of neutron and v - velocity

K.E = \frac{p^2}{2m} ...........................(1) \because v = \frac{p}{m}

p = \sqrt{2m K.E}

de Broglie wavelength \lambda = \frac{h}{p}

\lambda = \frac{h}{\sqrt{2 m_n K.E}} \because \text{ equation (1)}

Where, h - Planck's constant( 6.63* 10^{-34} ), m_n - mass of neutron

\lambda = \frac{6.638 10^{-34}}{2 * 1.675* 10^{-27} * 150 * 1.6* 10^{-19} }

\lambda = 2.327 * 10^{-12} m

Hence, de Broglie wavelength of neutron \lambda = 2.327 * 10^{-12} m

Step 2: Checking neutron energy is suitable for the crystal diffraction or not

Exercise 11.31:

Inter atomic spacing \lambda = 1 Angstrom = 10^{-10} m

de Broglie wavelength of neutron \lambda = 2.327 * 10^{-12} m

de Broglie wavelength of neutron < 100 * Inter atomic spacing

Therefore, a 150 eV energy neutron beam does not suit diffraction experiments.

Step 1: de Broglie wavelength neutron at room temperature

Neutron temperature T = 27^\degree C

T = 300 K \because 0\degree C = 273 K

Kinetic energy at temperature T

K.E = \frac{3}{2} kT

Where, k - Boltzmann constant = 1.38 * 10^{-23} J/mol * K

de Broglie wavelength neutron \lambda = \frac{h}{p} ...................................(1)

We know, momentum p = \sqrt{2m K.E}

Substituting p value in equation (1)

\lambda = \frac{h}{\sqrt{2m K.E}}

\lambda = \frac{h}{\sqrt{3 m kT}}

\lambda = \frac{6.63 * 10^{-34}}{3 * 1.675* 10^{-27} * 1.38 * 10^{-23} * 300 }

\lambda = 1.447 * 10^{-10} m

Hence, de Broglie wavelength neutron \lambda = 1.447 * 10^{-10} m

Step 2: Make sure the thermalised neutrons undergoes diffraction.

Broglie wavelength of thermalised neutron \lambda = 1.447 * 10^{-10} m

Inter atomic spacing \lambda = 1 Angstrom = 10^{-10} m

Broglie wavelength of thermalised neutron \approx Inter atomic spacing

Therefore, Before using the high-energy neutron beam for diffraction, it must first be thermalized.