Paragraph: Wave property of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see in figure).
Question: Electrons accelerated by potential $$V$$ are diffracted from a crystal. If $$d = 1\ \AA$$ and $$i = 30^\circ$$, $$V$$ should be about $$(h = 6.6 \times 10^{-34}\ Js,\ m_e = 9.1 \times 10^{-31}\ kg,\ e = 1.6 \times 10^{-19}\ C)$$

For electron diffraction from a crystal we need two relations: Bragg’s law for constructive interference and the de-Broglie relation for the wavelength of an electron accelerated through a potential $$V$$.

Step 1: Convert the given angle to the “glancing angle” used in Bragg’s law
Bragg’s law is $$2d\,\sin\theta = n\lambda$$, where $$\theta$$ is the glancing angle (angle between the incident beam and the crystal plane).
The question states that the electrons strike the crystal with an angle of incidence $$i = 30^\circ$$ measured from the normal to the plane. Hence the glancing angle is
$$\theta = 90^\circ - i = 90^\circ - 30^\circ = 60^\circ.$$
Therefore $$\sin\theta = \sin 60^\circ = \cos 30^\circ.$$ Using this in Bragg’s law for the first order $$n = 1$$:

$$\lambda = 2d\sin\theta = 2d\cos i.$$

With $$d = 1\ \text{\AA} = 1\times10^{-10}\ \text{m}$$ and $$\cos 30^\circ = \tfrac{\sqrt3}{2}$$, we get
$$\lambda = 2(1\ \text{\AA})\!\left(\tfrac{\sqrt3}{2}\right) = \sqrt3\ \text{\AA} \approx 1.732\ \text{\AA} = 1.732\times10^{-10}\ \text{m}.$$

Step 2: Relate the wavelength to the accelerating potential
For an electron accelerated through potential $$V$$,
$$\lambda = \frac{h}{\sqrt{2m_e e V}}.$$ Rearranging,

$$V = \frac{h^{2}}{2m_e e\,\lambda^{2}}.$$

Step 3: Substitute the numerical values
$$h = 6.6\times10^{-34}\ \text{J·s},\; m_e = 9.1\times10^{-31}\ \text{kg},\; e = 1.6\times10^{-19}\ \text{C},\; \lambda = 1.732\times10^{-10}\ \text{m}.$$ $$\begin{aligned} V &= \frac{(6.6\times10^{-34})^{2}} {2(9.1\times10^{-31})(1.6\times10^{-19})(1.732\times10^{-10})^{2}} \\[4pt] &= \frac{4.356\times10^{-67}} {8.736\times10^{-69}} \\[4pt] &\approx 50\ \text{V}. \end{aligned}$$

Hence the accelerating potential required is about $$50\ \text{V}$$.

Option B which is: 50 V