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(A) The de Broglie wavelength $\lambda$ of an electron is given by $\lambda = \frac{h}{mv} = \frac{h}{p}$, where $h$ is Planck's constant, $m$ is the

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(A) The de Broglie wavelength $\lambda$ of an electron is given by $\lambda = \frac{h}{mv} = \frac{h}{p}$, where $h$ is Planck's constant, $m$ is the mass, and $v$ is the velocity of the electron.For $\lambda$ to increase, the velocity $v$ of the electron must decrease.In case $(A)$, the electron (charge $-e$) moves in the direction of the electric field $\vec{E}$. The force on the electron is $\vec{F} = -e\vec{E}$, which acts opposite to its motion, causing the electron to decelerate (velocity decreases).In case $(B)$, the electron moves opposite to the electric field $\vec{E}$. The force $\vec{F} = -e\vec{E}$ acts in the direction of motion, causing the electron to accelerate (velocity increases).In cases $(C)$ and $(D)$, the magnetic force $\vec{F} = q(\vec{v} 	imes \vec{B})$ is perpendicular to the velocity, so the speed of the electron remains constant.Therefore, the velocity decreases only in case $(A)$, leading to an increase in the de Broglie wavelength.

The figure shows four situations in which an electron is moving in an electric or magnetic field. In which case is the de Broglie wavelength of the electron increasing?

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