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(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{mv}$.Case $(i)$: The electron is moving opposite to the electric field. Since the electr

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Decreasing,same

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(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{mv}$.Case $(i)$: The electron is moving opposite to the electric field. Since the electron is negatively charged,it experiences a force in the direction of the electric field,which is opposite to its velocity. This force acts as a retarding force,causing the velocity $v$ of the electron to decrease. As $v$ decreases,the de-Broglie wavelength $\lambda = \frac{h}{mv}$ increases.Case $(ii)$: The electron is moving perpendicular to a magnetic field. The magnetic force $\vec{F} = q(\vec{v} 	imes \vec{B})$ acts perpendicular to the velocity of the electron. $A$ magnetic force acting perpendicular to the velocity only changes the direction of motion,not the magnitude of the velocity $v$. Since the speed $v$ remains constant,the de-Broglie wavelength $\lambda = \frac{h}{mv}$ remains the same.Therefore,the wavelength increases in case $(i)$ and remains the same in case $(ii)$.

An electron is moving through a field. It is moving $(i)$ opposite to an electric field and $(ii)$ perpendicular to a magnetic field as shown. For each situation,determine the de-Broglie wavelength of the electron.

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