A beam of electrons of energy E scatters from a target having atomic spacing of 1A0. The first maximum intensity occurs at θ=600. Then Ein eV is _______(Planck constant h=6.64×10−34Js,1eV=1.6×10−19J,electron mass m=9.1×10−31kg)

Bragg's Condition for diffraction,2dsinθ=nλFor first maxima, n=1⇒2d× sin600=λ⇒d3=λ⇒λ=3A0Now, λ=h2mE ⇒E=h22mλ2=6.64×10−3422×9.1×10−31×3×10−102 ⇒E=0.8×10−17J ⇒E=0.8×10−171.6×10−19=50 eV

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A beam of electrons of energy $E$ scatters from a target having atomic spacing of $1 \overset{0}{A}$ . The first maximum intensity occurs at $θ = 60^{0}$ . Then $E (i n e V)$ is _______
(Planck constant $h = 6.64 \times 10^{- 34} J s, 1 e V = 1.6 \times 10^{- 19} J, e l e c t r o n m a s s m = 9.1 \times 10^{- 31} k g)$

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Detailed Solution

$\begin{array}{l} B r a g g' s C o n d i t i o n f o r d i f f r a c t i o n, \\ 2 d \sin θ = n λ \\ F o r f i r s t m a x i m a, n = 1 \\ \Rightarrow 2 d \times \sin 60^{0} = λ \\ \Rightarrow d \sqrt{3} = λ \\ \begin{array}{l} \Rightarrow λ = \sqrt{3} \overset{0}{A} \\ N o w, λ = \frac{h}{\sqrt{2 m E}} \\ \Rightarrow E = \frac{h^{2}}{2 m λ^{2}} = \frac{{(6.64 \times 10^{- 34})}^{2}}{2 \times 9.1 \times 10^{- 31} \times {(\sqrt{3} \times 10^{- 10})}^{2}} \end{array} \\ \begin{array}{l} \Rightarrow E = 0.8 \times 10^{- 17} J \\ \Rightarrow E = \frac{0.8 \times 10^{- 17}}{1.6 \times 10^{- 19}} = 50 e V \end{array} \end{array}$

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