In the Bragg scattering of a beam of electrons each of mass m and velocity v by a nickel crystal, the first maximum is observed at θ = 3 0 ∘ ( θ being the angle the beam makes with the crystal plane). What is the inter-planar distance d for the crystal?

For Bragg's scattering of a beam of electron, the inter-planar distance ' d ' is given by 2 d sin θ = n λ Here θ = 3 0 ∘ ⇒ sin 3 0 ∘ = 2 1 ; n = 1 ∴ d = 2 sin θ n λ = 2 × 2 1 1 × λ = λ or d = mv h (According to de-Broglie, wavelength associated with an electron λ = mv h )

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Physics - Wave Optics Question with Solution | TestHub

PhysicsWave OpticsDiffractionHard2 minPYQ_2022

PhysicsHardsingle choice

In the Bragg scattering of a beam of electrons each of mass $m$ and velocity $v$ by a nickel crystal, the first maximum is observed at $θ = 3 0^{\circ}$ ( $θ$ being the angle the beam makes with the crystal plane). What is the inter-planar distance $d$ for the crystal?

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Solution:

For Bragg's scattering of a beam of electron, the inter-planar distance ' $d$ ' is given by $2 d sin θ = n λ$ Here $θ = 3 0^{\circ}$ $\Rightarrow sin 3 0^{\circ} = \frac{1}{2}; n = 1 ∴ d = \frac{n λ}{2 sin θ} = \frac{1 \times λ}{2 \times \frac{1}{2}} = λ$ or $d = \frac{h}{mv}$ (According to de-Broglie, wavelength associated with an electron $λ = \frac{h}{mv}$ )

Stream:VITEEESubject:PhysicsTopic:Wave OpticsSubtopic:Diffraction

⏱ 2mℹ️ Source: PYQ_2022

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