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.

Question

Philoid · Accepted Answer

Given:

Kinetic energy of neutron, E = 150 eV = 150 × 1.6 × 10^-19

E = 2.4 × 10^-17 J

Mass of neutron, m_n = 1.675 × 10^-27 Kg

Kinetic energy of a particle is given by,

We can write,

mv = (2 × m × KE)^0.5

P = (2 × m × KE)^0.5 ...(1)

Where,

m = mass of electron

v = velocity of electron

p = momentum of particle

By putting the equation (1) in de- Broglie equation, we get,

…(2)

Now substituting the values in eq (2) we get,

λ = 2.327 × 10^-12m

A neutron cannot be used in diffraction experiment as lattice spaces are of the order of few 10^-10 m, whereas the wavelength of a neutron beam is 100 times smaller.

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.(mn = 1.675 × 10–27 kg)

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.
(mn = 1.675 × 10^–27 kg)