A beam of electrons of energy $E$ scatters from a target having atomic spacing of $1\ \text{Å}$. The first maximum intensity occurs at $\theta = 60^\circ$. Then $E$ (in eV) is ______. (Planck's constant $h = 6.64\times10^{-34}$ J s, $1\ \text{eV} = 1.6\times10^{-19}$ J, electron mass $m = 9.1\times10^{-31}$ kg)
Numerical value type. Enter your answer.
Answer: 50
Treating the scattering as Bragg reflection from planes of spacing $d = 1$ Å, the first maximum gives $2d\sin\theta = \lambda$:
$$\lambda = 2\times1\times\sin60^\circ = \sqrt3\ \text{Å}$$
$$E = \frac{h^{2}}{2m\lambda^{2}} = \frac{(6.64\times10^{-34})^{2}}{2\times9.1\times10^{-31}\times3\times10^{-20}} \approx 8.1\times10^{-18}\ \text{J} \approx 50\ \text{eV}$$
Solution by Sreeraj P, M.Sc Physics