Answer

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**Hint:**The wave associated with moving particles is called matter wave or de-Broglie wave whose wavelength is called de-Broglie wavelength.

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

Where h is planck's constant, m is mass of particle and v is velocity of particle.

The kinetic energy is given by

$ \begin{align}

& v=\sqrt{\dfrac{2K}{m}} \\

& \lambda =\dfrac{h}{m\sqrt{2k/m}}=\dfrac{h}{\sqrt{2km}} \\

\end{align} $

Using the above formula, we will get the required result.

**Complete step by step solution:**

We have given, an electron having kinetic energy = 100 eV

An electron is accelerated through a potential difference, So additional kinetic energy is gains $ =\text{e}\left( 50\text{V} \right)=50\text{ev} $

Total kinetic energy is equal to $ =50 $ ev $ +100 $ ev $ =150 $ ev

$ =150\times 1\cdot 6\times {{10}^{-19}} $ J

The de-Broglie wavelength can be calculated from the following formula.

$ \begin{align}

& \lambda =\dfrac{h}{mv} \\

& \lambda =\dfrac{h}{\sqrt{2mk}} \\

\end{align} $ --- (1)

Mass of electron = $ 9.11\times {{10}^{-31}}kg $

Planck's constant = $ 6.63\times {{10}^{-34}} $ Js

Put all the values in eq. (1)

$ \begin{align}

& \lambda =\dfrac{6.63\times {{10}^{-34}}}{\sqrt{2\times 9.11\times {{10}^{-31}}\times 150\times 1.6\times {{10}^{-19}}}} \\

& \lambda =\dfrac{6.63\times {{10}^{-34}}}{\sqrt{4372.8\times {{10}^{-50}}}} \\

& \lambda =\dfrac{6.63\times {{10}^{-34}}}{66.13\times {{10}^{-25}}} \\

& \lambda =\dfrac{6.63\times {{10}^{-34}}}{6.613\times {{10}^{-24}}} \\

& \lambda ={{10}^{-10}}m \\

\end{align} $

$ \lambda =1\text{A}{}^\circ $

Hence, the de-Broglie wavelength is $ 1\text{A}{}^\circ $ .

**So, the correct option is “A”.**

**Note:**

Significance of de-Broglie equation.

De-Broglie proposed that as light exhibits both wave-like and particle-like properties, matter exhibits wave-like and particle-like properties. This nature was described as dual behavior of matter.

Another method.

Use direct formula, $ \lambda =\dfrac{12.27}{\sqrt{v}} $

V can be calculated as $ \dfrac{1}{2}m{{v}^{2}}=eV $ , Put the values in the above formula, we will get the same result.

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