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(D) The de Broglie wavelength $\lambda$ of a particle with kinetic energy $E$ and mass $m$ is given by the formula $\lambda = \frac{h}{\sqrt{2mE}}$.Si

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An electron

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(D) The de Broglie wavelength $\lambda$ of a particle with kinetic energy $E$ and mass $m$ is given by the formula $\lambda = \frac{h}{\sqrt{2mE}}$.Since the kinetic energy $E$ and Planck's constant $h$ are the same for all particles,we have the relation $\lambda \propto \frac{1}{\sqrt{m}}$.This implies that the particle with the smallest mass $m$ will have the largest de Broglie wavelength $\lambda$.Comparing the masses of the given particles: $m_{	ext{electron}} Since the electron has the smallest mass,it will have the largest de Broglie wavelength.

Of the following particles having the same kinetic energy,the one which has the largest de Broglie wavelength is

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