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What is the de Broglie wavelength of the electron accelerated through a potential difference of $100\,V$ ?
A) $12.27\,\mathop A\limits^0 $
B) $1.227\,\mathop A\limits^0 $
C) $0.1227\,\mathop A\limits^0 $
D) $0.001227\,\mathop A\limits^0 $

Last updated date: 17th Jun 2024
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Hint:The De Broglie frequency is a frequency that is shown in all the objects in quantum mechanics which decides the likelihood density of finding the object at a given purpose of the setup space. The de Broglie frequency of a molecule and its relative to its momentum are contrary. Use the formula of the de Broglie wavelength to find the answer.

Complete step-by-step solution:
The de Broglie wavelength for macroscopic objects is given as
$\lambda = \dfrac{h}{p}$
Here, $\lambda $ is the de Broglie wavelength for the given particle

$h$ is constant having value $h = 6.63 \times {10^{ - 24}}$
And $p$ is the momentum of the particle
Now, we know that momentum and the potential difference are related as follows:
$p = \sqrt {2{m_e}V} $
Here, $m$ is the mass of the particle (electron, in our case)

$V$ is the potential difference.
Substituting the value of momentum in above equation, we get
$\lambda = \dfrac{h}{{\sqrt {2{m_e}V} }}$
Substituting the values, we get
$ \Rightarrow \lambda = \dfrac{{6.63 \times {{10}^{ - 24}}}}{{\sqrt {2 \times 9.1 \times {{10}^{ - 31}} \times 100} }}$
$ \Rightarrow \lambda = 1.227 \times {10^{ - 10}}\,m$
$ \therefore \lambda = 1.227\mathop A\limits^0 \,m$

Thus, option B is the correct option.

Additional details:The de Broglie frequency carries on the photon frequency with the related momentum, which joins particles and waves, de Broglie frequencies are viewed as likelihood waves related to the wave work. It is accepted, in quantum mechanics, that the squared sufficiency of the wave work at a given point in the facilitated portrayal decides the likelihood density of finding the molecule now.

Note:Electromagnetic potential of particles diminishes in the reverse extent of the good ways from the molecule to the perception point, the potential of solid collaboration in the gravitational examples of solid communication acts in a similar way. Remember the relation between potential difference and momentum.