An electron beam can undergo diffraction by crystals. Through what potential should a beam of electrons be accelerated so that its wavelength becomes equal to $1.54\;\mathop {\rm A}\limits^ \circ $?
A) $13.6\;volts$
B) $1.54\;volts$
C) $1\;volt$
D) $63.44\;volts$

Hint: An electron beam undergoes diffraction by crystal, then the electrons in the beam are accelerated by the crystal. Thus, after acceleration the kinetic energy of the electron is the same as the electron volt of the electron. Using the wavelength of the electron, the kinetic energy of the electron can be calculated with the help of deBroglie wavelength relation. By equating the kinetic energy and electron volt, the solution of potential of electron can be obtained.

Formulae used:
The kinetic energy of the electron is given by,
$K.E = \dfrac{1}{2}m{v^2}$
Where,$K.E$ is the kinetic energy of the electron, $m$ is the mass of the electron and $v$ is the velocity of the electron.

Electron volt of electron,
$E.V = qV$
Where, $E.V$ is electron volt, $q$ is the charge of an electron and $V$ is the potential energy of an electron.
DeBroglie wavelength is given by,
$\lambda = \dfrac{h}{{mv}}$
Where, $\lambda $ is the wavelength, $h$ is the Planck’s constant, $m$ is the mass of the electron and $v$ is the velocity of the electron.

Complete step by step solution:
Mass of the electron, $m = 9.108 \times {10^{ - 31}}\;kg$
Planck’s constant, $h = 6.626 \times {10^{ - 34}}\;{m^2}kg{s^{ - 1}}$
Charge of electron, $q = 1.602 \times {10^{ - 19}}\;C$
Wavelength of electron, $\lambda = 1.54\;\mathop {\rm A}\limits^ \circ $ (or) $\lambda = 1.54 \times {10^{ - 10}}\;m$


Since, after acceleration the kinetic energy of the electron is as same as the electron volt of the electron,
$K.E = E.V$
Substitute the values of $K.E$ and $E.V$in the above equation,
$
  \dfrac{1}{2}m{v^2} = qV \\
  V = \dfrac{{m{v^2}}}{{2q}}\;.......................................\left( 1 \right) \\
 $
By using DeBroglie wavelength,
$\lambda = \dfrac{h}{{mv}}$
Rearrange the above equation,
$v = \dfrac{h}{{m \times \lambda }}$
Substitute the value of $v$ in equation (1),
$
  V = \dfrac{{m \times {{\left( {\dfrac{h}{{m \times \lambda }}} \right)}^2}}}{{2 \times q}} \\
  V = \dfrac{{m \times \left( {\dfrac{{{h^2}}}{{{m^2} \times {\lambda ^2}}}} \right)}}{{2 \times q}} \\
  V = \dfrac{{{h^2}}}{{2 \times q \times m \times {\lambda ^2}}}\;.....................................\left( 2 \right) \\
 $
Substitute the values of $h$, $q$, $m$ and $\lambda $ in the equation (2),
$
  V = \dfrac{{{{\left( {6.626 \times {{10}^{ - 34}}\;{m^2}kg{s^{ - 1}}} \right)}^2}}}{{2 \times \left( {1.602 \times {{10}^{ - 19}}\;C} \right) \times \left( {9.108 \times {{10}^{ - 31}}\;kg} \right) \times {{\left( {1.54 \times {{10}^{ - 10}}\;m} \right)}^2}}} \\
  V = \dfrac{{4.3903876 \times {{10}^{ - 67}}\;{m^4}k{g^2}{s^{ - 2}}}}{{2 \times 1.602 \times {{10}^{ - 19}}\;C \times 9.108 \times {{10}^{ - 31}}\;kg \times 2.71 \times {{10}^{ - 20}}\;{m^2}}} \\
  V = 63.57\;V \\
  V \simeq 63.44\;volts \\
 $

Hence, the option (D) is correct.

Note: In the given question, the electron is diffracted by the crystals. Then the electron is accelerated by diffraction with a certain wavelength and some kinetic energy. By the data collected from the electron beam, the properties of the crystal are also calculated. These electron diffractions are also used in transmission electron microscopes (TEM), to study the properties of materials.