An electron beam can undergo diffraction by crystals. if \[63.573{\text{ V}}\] potential give a beam of electrons to be accelerated so that its Wavelength becomes equal to:

Hint:You can use the rearranged form of the De-broglie equation as \[u = \dfrac{h}{{m \times \lambda }}\]
You can also use the relationship between the potential energy and kinetic energy as \[e \times V = \dfrac{{m \times {u^2}}}{2}\]

Complete step-by-step solution:
Let the wavelength become equal to \[x{\text{ }}{{\text{A}}^o}\].
According to the De-Broglie hypothesis, the relationship between the wavelength \[\lambda \] of an electron and its momentum ‘p’ is given by the following expression.
\[\lambda = \dfrac{h}{P} = \dfrac{h}{{m \times u}}\]
Here, ‘h’ is Planck's constant, m is the mass of the electron and ‘u’ is its velocity.
Rearrange the above expression
\[u = \dfrac{h}{{m \times \lambda }}\]… …(1)

When an electron beam is accelerated through a potential of V, then its potential energy is given by the expression \[P.E = e \times V\]
The kinetic energy of the electron is given by the expression \[K.E = \dfrac{{m \times {u^2}}}{2}\]
But, the magnitude of the potential energy is equal to the magnitude of kinetic energy. Hence,
\[e \times V = \dfrac{{m \times {u^2}}}{2}\]
 Rearrange the above equation
\[V = \dfrac{{m \times {u^2}}}{{2 \times e}}\]… …(2)
Substitute the value of the speed ‘u’ from equation (1) into equation (2)
\[
  V = \dfrac{{m \times {{\left( {\dfrac{h}{{m \times \lambda }}} \right)}^2}}}{{2 \times e}} \\
  V = \dfrac{{{h^2}}}{{2 \times e \times m \times {\lambda ^2}}} \\
 \]
Substitute values in the above expression
\[
  V = \dfrac{{{h^2}}}{{2 \times e \times m \times {\lambda ^2}}} \\
  63.573{\text{ V}} = \dfrac{{{{\left( {6.626 \times {{10}^{ - 34}}{\text{ Js}}} \right)}^2}}}{{2 \times 1.6 \times {{10}^{ - 19}} \times 9.1 \times {{10}^{ - 31}}{\text{ kg}} \times {{\left( {x \times {{10}^{ - 10}}{\text{m}}} \right)}^2}}} \\
  x = 1.54{\text{ }}{{\text{A}}^o} \\
 \]

Hence, the associated wavelength becomes equal to \[1.54{\text{ }}{{\text{A}}^o}\] .

Note:According to the De-Broglie hypothesis, the matter has dual nature. Matter can also show wave-like properties. The electron diffraction shows the wave-like behaviour of the electron.
 Do not substitute the value of mass of the electron in the unit of grams as it will lead to an error. Also use conversion factor for conversion of unit of wavelength from meters to angstroms.