Question

An electron beam of energy 10 KeV is incident on metallic foil. If the interatomic distance is 0.55Å . Find
the angle of diffraction.

Answer 1

Answer:

Angle of Diffraction is 12.89° approximately.

Explanation:

Given,

An electron beam of energy 10 KeV is incident on metallic foil.

The interatomic distance (D) is 0.55Å.

To find : the angle of diffraction (ϕ).

Solution:

The relation between interatomic distance and wavelength is:

λ = D sinϕ

And, Relation between wavelength and Energy is:

λ = 12.27Å / √V

So,  D sin ϕ = 12.27Å /√V

or, 0.55 * sinϕ = 12.27*10⁻¹⁰ / √(10 * 10³)

or, sin ϕ = 12.27 / (0.53 * 100)

or, sin ϕ = 0.2231

or, ϕ = sin⁻¹ (0.2231)

or, ϕ ≈ 12.89°

Therefore, Angle of Diffraction is 12.89° approximately.

Answer 2

Given:

• Energy of electron beam = 10 KeV.
• Interatomic distance = 0.55 Å.

To Find:

• The angle of diffraction.

$\;\large\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

We are given the energy of an electron beam and Interatomic distance in the question. We need to find the angle of diffraction [θ = ?]. To find the diffraction of angle here, we need to apply the formula of de broglie's wavelength and Bragg's law.

Solution:

To find the angle of diffraction = ?

First, convert the energy from KeV to Joules.

$\sf E = 10 KeV = 10 \times 1000 \times 1.6 \times 10^{-19} J$

De broglie's wavelength = λ = $\sf \dfrac{h}{\sqrt{2mE}}$

$\sf \dfrac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 10 \times 1000 \times 1.6 \times 10^{-19}}}$

$\dfrac{\dfrac{663}{100} \times 10^{-34}}{\sqrt{2 \times \dfrac{91}{10} \times 10^{-31} \times 10 \times 10^3 \times \dfrac{8}{5} \times 10^{-19}}}$

$\sf \dfrac{\dfrac{663}{10^2} \times 10^{-34}}{\sqrt{2 \times 91 \times 10^{-31} \times 10^3 \times \dfrac{8}{5} \times 10^{-19}}}$

$\sf \dfrac{663}{10^{36} \times \sqrt{2 \times 91 \times 10^{-31} \times 10^{3} \times \dfrac{8}{5} \times 10^{-19}}}$

$\sf \dfrac{663}{10^{36} \times 10 \times 2 \sqrt{728 \times 10^{-50}}}$

$\sf \dfrac{663}{4 \times 10^{12} \sqrt{182}}$

$\sf \dfrac{51 \sqrt{182}}{56 \times 10^{12}}$

$\large{\underline{\boxed{\sf 1.229 \times 10^{-11} m}}}$

Now,

Bragg's law 2nd, sinθ = nλ

First order diffraction, n = 1

Here,

$\implies \sf sin \theta = \dfrac{n \lambda}{2d}$

$\implies \sf \dfrac{1.229 \times 10^{-11}}{2 \times 0.55 \times 10^{-10}}$

$\implies \sf 0.1117$

$\large{\underline{\boxed{\sf \therefore \ \theta = 6.415^{\circ}}}}$