The equation $\lambda=\frac{1.227}{x} \mathrm{~nm}$ can be used to find the de-Brogli wavelength of an electron. In this equation $x$ stands for :

Where

$\mathrm{m}=$ mass of electron

$\mathrm{P}=$ momentum of electron

$\mathrm{K}=$ Kinetic energy of electron

$\mathrm{V}=$ Accelerating potential in volts for electron

<p>The de Broglie wavelength of a particle can be expressed as:</p> <p>$\lambda = \frac{h}{p}$</p> <p>where:</p> <ul> <li>$h$ is Planck&#39;s constant, and</li> <li>$p$ is the momentum of the particle.</li> </ul> <p>For an electron accelerated through a potential difference of $V$ volts, its kinetic energy $K$ is given by:</p> <p>$K = eV$</p> <p>where $e$ is the charge of an electron.</p> <p>The electron&#39;s momentum can be expressed in terms of its kinetic energy as:</p> <p>$p = \sqrt{2mK}$</p> <p>where $m$ is the mass of the electron.</p> <p>Substituting this into the de Broglie wavelength equation, we get:</p> <p>$\lambda = \frac{h}{\sqrt{2mK}}$</p> <p>Comparing this with the given equation $\lambda = \frac{1.227}{x} \, nm$, we see that $x$ must correspond to:</p> <p>$x = \sqrt{2mK}$</p> <p>Since $K = eV$, we can substitute this into our expression for $x$ to get:</p> <p>$x = \sqrt{2meV} = \sqrt{V}$</p> <p>where we&#39;ve used the fact that the mass and charge of an electron are constants.</p>