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# An electron with initial kinetic energy of 100 eV is accelerated through a potential difference of 50 V. Then find the de-Broglie wavelength of electron becomes (A) $1\text{A}{}^\circ$ (B) $\sqrt{1.5}\text{A}{}^\circ$ (C) $\sqrt{3}\text{A}{}^\circ$ (D) $12\cdot 27\text{A}{}^\circ$

Last updated date: 20th Jun 2024
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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.