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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 $

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Last updated date: 20th Jun 2024
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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.