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**Hint:**Remember that, when an electron is accelerated through a potential difference, the work done will be in the form of the kinetic energy of the electron, and the wavelength changes with the potential difference, thus wavelength can be calculated.

**Complete step by step solution:**

Let’s define all the terms described in the question. It is said that the potential difference through which the electron is accelerated is $100V$. Here in the question, we are asked to find the de Broglie wavelength associated with it.

As there is an acceleration, there is a motion of the electron. For the motion of electrons there should be some amount of work done .

Work done is given by charge multiplied by the potential difference,

That is, Work done, $W = q \times {V_0}$…………… (1)

That is, $W = K.E.$

$ \Rightarrow W = \dfrac{1}{2}m{v^2}$

Here we are multiplying both denominator and numerator with $m$ for the further calculations, we get,

$ \Rightarrow W = \dfrac{1}{2}m{v^2}$

$ \Rightarrow W = \dfrac{1}{2}m{v^2} \times \dfrac{m}{m}$

$ \Rightarrow W = \dfrac{{{{(mv)}^2}}}{{2m}}$……………….. (2)

We know the momentum, $P = mv$

So equation (2) will become as given below,

$ \Rightarrow W = \dfrac{{{P^2}}}{{2m}}$…………………(3)

Comparing the two equations (1) and (2), we get,

$P = \sqrt {2mq{V_0}} $

We know the de Broglie wavelength is given by the equation,

$\lambda = \dfrac{h}{P}$

Where, $h$ is the Planck's constant

Applying the value of the momentum and all other terms in this equation, we will get,

$ \Rightarrow \lambda = \dfrac{h}{{\sqrt {2mq{V_0}} }}$

$ \Rightarrow \lambda = \dfrac{{6.63 \times {{10}^{ - 34}}}}{{\sqrt {2 \times 9.1 \times {{10}^{ - 31}} \times 1.6 \times {{10}^{ - 19}} \times 100} }}$

$ \Rightarrow \lambda = \dfrac{{12.27}}{{\sqrt {{V_0}} }}$

$ \Rightarrow \lambda = \dfrac{{12.27}}{{\sqrt {100} }}$

\[ \Rightarrow \lambda = 1.227\mathop A\limits^0 \]

That is the de Broglie wavelength of the electron when it is accelerated through a potential difference of $100V$, \[\lambda = 1.227\mathop A\limits^0 \]

**This value of wavelength belongs to the X-ray part of the electromagnetic spectrum.**

**Note:**The electromagnetic (EM) spectrum is the range of all types of electromagnetic radiation. The types of EM radiation that make up the electromagnetic spectrum are radio, microwaves, infrared light, visible, ultraviolet light, X-rays and gamma-rays.

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