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(A) The electron enters the region between the plates with an initial kinetic energy of $200 \, eV$.Since the plates are connected to a $100 \, V$ pow

Vedclass · Accepted Answer

$1.22 \, \mathring{A}$

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(A) The electron enters the region between the plates with an initial kinetic energy of $200 \, eV$.Since the plates are connected to a $100 \, V$ power supply, there is an electric field between them. Based on the diagram, the electron moves against the electric field (from positive to negative plate), which acts as a retarding potential.Therefore, the final kinetic energy of the electron when it emerges from the second plate is $K_f = 200 \, eV - 100 \, eV = 100 \, eV$.The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{12.27}{\sqrt{V}} \, \mathring{A}$, where $V$ is the kinetic energy in $eV$.Substituting $V = 100 \, eV$:$\lambda = \frac{12.27}{\sqrt{100}} = \frac{12.27}{10} = 1.227 \, \mathring{A} \approx 1.22 \, \mathring{A}$.

Two large parallel plates are connected to a $100 \, V$ power supply. These plates have a fine hole at the centre. An electron having energy $200 \, eV$ is directed such that it passes through the holes. When it emerges, its de-Broglie wavelength is ............. $\mathring{A}$

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