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(D) The electron enters the region between the plates with an initial kinetic energy of $200\, eV$.The electric field between the plates is directed f

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None of these

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(D) The electron enters the region between the plates with an initial kinetic energy of $200\, eV$.The electric field between the plates is directed from the positive plate to the negative plate. Since the electron is negatively charged,it experiences a force opposite to the electric field,i.e.,towards the positive plate.As the electron moves from the first plate to the second plate,it moves against the electric field,so it loses kinetic energy.The potential difference between the plates is $V = 100\, V$. The work done by the electric field on the electron is $W = -e 	imes V = -100\, eV$.The final kinetic energy of the electron when it emerges from the second plate is $K_f = K_i - 100\, eV = 200\, eV - 100\, eV = 100\, eV$.The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{12.27}{\sqrt{V_{eff}}}\, \mathring{A}$,where $V_{eff}$ is the kinetic energy in $eV$.$\lambda = \frac{12.27}{\sqrt{100}} = \frac{12.27}{10} = 1.227\, \mathring{A}$.Since $1.227\, \mathring{A}$ is approximately $1.23\, \mathring{A}$,and this value is not explicitly listed in the options $A, B, C$,the correct choice is $D$.

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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