(A) What is de Broglie hypothesis?
(B) Write the formula for de Broglie wavelength.
(C) Calculate de Broglie wavelength associated with an electron accelerated by a potential difference of $100\,volts$.
Given the mass of the electron = $9.1 \times {10^{ - 31}}kg$, $h = 6.634 \times {10^{ - 34}}Js$, $1\,eV = 1.6 \times {10^{ - 19}}J$

Hint De Broglie's hypothesis related the wavelength of a particle to its momentum. In quantum mechanics, as we zoom into the subatomic level, matter doesn’t always necessarily exhibit the particle behavior as we expect. It also behaves as a wave at the microscopic level.
Formulas used: For de Broglie wavelength $\lambda = \dfrac{h}{p}$ and momentum $p = \sqrt {2meV} $

Complete Step by step solution
The de Broglie hypothesis proposes that all matter exists in a dual state of waves and particles exhibiting wave-like properties. It relates the observed wavelength of a particle to its momentum.
The de Broglie wavelength $\lambda $ of a particle is associated with its momentum $p$, as per the formula $\lambda = \dfrac{h}{p}$, where $h$ is the Planck constant.
The momentum $p$ in the above formula is given by $p = \sqrt {2meV} $, where $m$ is the mass of the electron, $e$ is the charge of an electron, and $V$ is the potential difference through which the electron has been accelerated.
We calculate the de Broglie wavelength of an electron been accelerated by a potential difference of $100\,volts$ using the formula $\lambda = \dfrac{h}{p}$, and substituting the momentum $p = \sqrt {2meV} $.
Thus we get $\lambda = \dfrac{h}{{\sqrt {2meV} }}$
Substituting the values of $h$ ,$m$ and $e$ we get
$\lambda = \dfrac{{6.634 \times {{10}^{ - 34}}Js}}{{\sqrt {2 \times 9.1 \times {{10}^{ - 31}}kg \times 1.6 \times {{10}^{ - 19}}C \times 100\,V} }}$
Upon further solving we get,
$\lambda = \dfrac{{6.634 \times {{10}^{ - 10}}}}{{\sqrt {2 \times 9.1 \times 1.6} }}m$
$ \Rightarrow \lambda = \dfrac{{6.634 \times {{10}^{ - 10}}}}{{5.396}}m = 1.229 \times {10^{ - 10}}\,m$

Hence the required answer is $1.23 \times {10^{ - 10}}\,m$.

Note The mass of the electron used in this formula is the rest mass of the particle and we generally use the rest mass in such equations unless specified. Also, the charge of the electron has been replaced in the formula with $1\,eV$, as an electron volt is the amount of energy gained by an electron when it is accelerated in a potential difference of $1\,Volt$ (numerical values of both being the same).