In Davisson and Germer experiment if the peak diffraction pattern is obtained at the voltage of 54 V, find the wavelength associated with the electron used for obtaining the diffraction pattern. What should be the angle corresponding to this voltage, which gives maximum diffraction.

Solution

Given,
Voltage, V = 54 V

Using the formula,
$λ = \frac{1.227}{\sqrt{V}} = \frac{1.227}{\sqrt{54}} = 0.165 nm = 1.65 Å$

From Bragg's law, $2 d \sin θ = nλ = λ$ [For first order diffraction, n = 1]

For Ni crystal,
Inter-atomic spacing, $d = 0.91 Å$
Substituting this in the formula,
We have,
$2 \times 0.91 \times 10^{- 10} \sin θ = 1.65 \times 10^{- 10}$

$θ = 65 °$
But, $θ = \frac{1}{2} (180 ° - ϕ)$
Hence,
for $θ = 65 °, the scattering angle, ϕ = 50 ° .$

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Wave Optics

In Davisson and Germer experiment if the peak diffraction pattern is obtained at the voltage of 54 V, find the wavelength associated with the electron used for obtaining the diffraction pattern. What should be the angle corresponding to this voltage, which gives maximum diffraction.

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