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(D) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.Since the magnetic force $\vec{F} = q(\vec{v} 	imes \vec{B})$ is alw

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$\lambda_0$

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(D) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.Since the magnetic force $\vec{F} = q(\vec{v} 	imes \vec{B})$ is always perpendicular to the velocity $\vec{v}$,the magnetic field does no work on the electron.According to the work-energy theorem,the kinetic energy of the electron remains constant.Since the kinetic energy $K = \frac{1}{2}mv^2$ is constant,the speed $v$ of the electron remains constant.As the speed $v$ remains constant,the momentum $p = mv$ remains constant.Therefore,the de Broglie wavelength $\lambda = \frac{h}{p}$ remains constant over time.Thus,the wavelength at time $t$ is equal to the initial wavelength $\lambda_0$.

An electron with mass $m$ and an initial velocity $(t=0)$ $\vec{v} = v_0 \hat{i}$ $(v_0 > 0)$ enters a magnetic field $\vec{B} = B_0 \hat{j}$. If the initial de Broglie wavelength at $t=0$ is $\lambda_0$,then its value after time $t$ would be:

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