Question

A beam of $ \mathbf{5 0 - k e V} $ electrons is directed at a crystal and diffracted electrons are found at an angle of $ 50^{\circ} $ relative to the original beam. What is the spacing of the atomic planes of the crystal? Is relativistic calculation needed for $ \lambda $. 3-26: As in the example presented in the text in Section 3.5, a beam deflection of $ 50^{\circ} $ corresponds to a scattering angle of $ \frac{1}{2}\left(180^{\circ}-50^{\circ}\right)=65^{\circ} $. The formula obtained from Problem 3-10 gives \[ \begin{aligned} \lambda & =\frac{h c}{\sqrt{\mathrm{KE}\left(\mathrm{KE}+2 m c^{2}\right)}} \\ & =\frac{1.240 \times 10^{-6} \mathrm{eV} \cdot \mathrm{~m}}{\sqrt{50 \mathrm{eV}\left(50+2\left(511 \times 10^{3}\right)\right)} \times 10^{3} \mathrm{eV}}=5.356 \times 10^{-12} \mathrm{~m} \end{aligned} \] keeping extra significant figures. The spacing of the planes in the crystal is then, using Equation (2.13), \[ d=\frac{\lambda}{2 \sin \theta}=\frac{5.356 \times 10^{-12} \mathrm{~m}}{2 \sin \left(65^{\circ}\right)}=2.95 \times 10^{-12} \mathrm{~m}=2.95 \mathrm{pm} \] To the two significant figures given in the problem, the spacing is 3.0 pm .

          A beam of \( \mathbf{5 0 - k e V} \) electrons is directed at a crystal and diffracted electrons are found at an angle of \( 50^{\circ} \) relative to the original beam. What is the spacing of the atomic planes of the crystal? Is relativistic calculation needed for \( \lambda \).

3-26: As in the example presented in the text in Section 3.5, a beam deflection of \( 50^{\circ} \) corresponds to a scattering angle of \( \frac{1}{2}\left(180^{\circ}-50^{\circ}\right)=65^{\circ} \). The formula obtained from Problem 3-10 gives
\[
\begin{aligned}
\lambda & =\frac{h c}{\sqrt{\mathrm{KE}\left(\mathrm{KE}+2 m c^{2}\right)}} \\
& =\frac{1.240 \times 10^{-6} \mathrm{eV} \cdot \mathrm{~m}}{\sqrt{50 \mathrm{eV}\left(50+2\left(511 \times 10^{3}\right)\right)} \times 10^{3} \mathrm{eV}}=5.356 \times 10^{-12} \mathrm{~m}
\end{aligned}
\]
keeping extra significant figures. The spacing of the planes in the crystal is then, using Equation (2.13),
\[
d=\frac{\lambda}{2 \sin \theta}=\frac{5.356 \times 10^{-12} \mathrm{~m}}{2 \sin \left(65^{\circ}\right)}=2.95 \times 10^{-12} \mathrm{~m}=2.95 \mathrm{pm}
\]

To the two significant figures given in the problem, the spacing is 3.0 pm .

$A beam of 5 0 - 𝐤 𝐞 𝐕 electrons is directed at a crystal and diffracted electrons are found at an angle of 50^∘ relative to the original beam. What is the spacing of the atomic planes of the crystal? Is relativistic calculation needed for λ. 3-26: As in the example presented in the text in Section 3.5, a beam deflection of 50^∘ corresponds to a scattering angle of (1)/(2)(180^∘-50^∘)=65^∘. The formula obtained from Problem 3-10 gives λ =(h c)/(√(KE(KE+2 m c^2))) =(1.240 × 10^-6eV· m)/(√(50 eV(50+2(511 × 10^3)))× 10^3eV)=5.356 × 10^-12 m keeping extra significant figures. The spacing of the planes in the crystal is then, using Equation (2.13), d=(λ)/(2 sinθ)=(5.356 × 10^-12 m)/(2 sin(65^∘))=2.95 × 10^-12 m=2.95 pm To the two significant figures given in the problem, the spacing is 3.0 pm .$

Added by Lisa B.

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