What is de Broglie wavelength of an electron has been accelerated through potential difference of 10000v?

In 1924 Louis de Broglie theorized that not only light posesses both wave and particle properties, but rather particles with mass - such as electrons - do as well. The wavelength of these 'material waves' - also known as the de Broglie wavelength - can be calculated from Planks constant $h$ divided by the momentum $p$ of the particle. In the case of electrons that is $$\lambda_{\text{de Broglie}} = \frac {h}{p_\text e}=\frac {h}{m_\text e\cdot v_\text e}$$ The acceleration of electrons in an electron beam gun with the acceleration voltage $V_{\rm{a}}$ results in the corresponding de Broglie wavelength $$\bbox[8px,border:2px solid red]{\lambda_{\text{de Broglie}} = \frac {h}{m_\text e\cdot \sqrt{2\cdot \frac{e}{m_\text e}\cdot V_{\text a}}}=\frac {h}{\sqrt{2\cdot m_\text e \cdot e\cdot V_{\text a}}}}$$ Proof of the de Broglie hypothesis will be experimentally demonstrated with the help of an electron diffraction tube on the following pages.

For electons with acceleration voltage $V_{\rm{a}}$, the following constansts will be used to calculate the de Broglie wavelengths:
$h=6{.}6\cdot 10^{-34}\,\text J\cdot \text s$, $ m_{\text e}=9{.}1\cdot 10^{-31}\,\text{kg}$, $e=1{.}6\cdot 10^{-19}\,{\text C}$

Note: The input only allows an acceleration voltage up to $V_{\rm{a}}=100000\,\rm{V}$. For higher acceleration voltage relativistic calculation should be used.

An electron is accelerated through a potential difference of 10000 V. The de Broglie wavelength associated with electron will be

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Solution : Here, `V=100` Volts.The de- Broglie wavelength `lambda " is " lambda =(1.227)/(sqrt(V))nm`. <br> `=(1.227)/(sqrt(100))=(1.227)/(10) =0.1227=0.123 nm`. <br> The value of de Broglie wavelength is associated with the wavelength of X-ray.