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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 \(\theta=30^{\circ}\) (\(\theta\) being the angle the beam makes with the crystal plane). What is the inter-planar distance \(d\) for the crystal?
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Verified Answer
The correct answer is:
\(\frac{\mathrm{h}}{\mathrm{mv}}\)
For Bragg's scattering of a beam of electron, the inter-planar distance ' \(\mathrm{d}\) ' is given by \(2 \mathrm{~d} \sin \theta=\mathrm{n} \lambda\)
Here \(\theta=30^{\circ}\)
\(\begin{aligned}
& \Rightarrow \sin 30^{\circ}=\frac{1}{2} ; \mathrm{n}=1 \\
& \therefore \mathrm{d}=\frac{\mathrm{n} \lambda}{2 \sin \theta}=\frac{1 \times \lambda}{2 \times \frac{1}{2}}=\lambda
\end{aligned}\)
or \(\mathrm{d}=\frac{\mathrm{h}}{\mathrm{mv}}\)
(According to de-Broglie, wavelength associated with an electron \(\lambda=\frac{\mathrm{h}}{\mathrm{mv}}\))
Here \(\theta=30^{\circ}\)
\(\begin{aligned}
& \Rightarrow \sin 30^{\circ}=\frac{1}{2} ; \mathrm{n}=1 \\
& \therefore \mathrm{d}=\frac{\mathrm{n} \lambda}{2 \sin \theta}=\frac{1 \times \lambda}{2 \times \frac{1}{2}}=\lambda
\end{aligned}\)
or \(\mathrm{d}=\frac{\mathrm{h}}{\mathrm{mv}}\)
(According to de-Broglie, wavelength associated with an electron \(\lambda=\frac{\mathrm{h}}{\mathrm{mv}}\))
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