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(C) The de Broglie wavelength $\lambda$ of a particle of mass $m$ accelerated through a potential difference $V$ is given by the formula:$\lambda = \f

Vedclass · Accepted Answer

$\lambda_e > \lambda_p$

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(C) The de Broglie wavelength $\lambda$ of a particle of mass $m$ accelerated through a potential difference $V$ is given by the formula:$\lambda = \frac{h}{\sqrt{2meV}}$where $h$ is Planck's constant and $e$ is the charge of the particle.Since both the proton and the electron are accelerated by the same potential difference $V$ and have the same magnitude of charge $e$,the wavelength is inversely proportional to the square root of the mass of the particle:$\lambda \propto \frac{1}{\sqrt{m}}$We know that the mass of an electron $(m_e)$ is significantly less than the mass of a proton $(m_p)$,i.e.,$m_e Therefore,$\frac{1}{\sqrt{m_e}} > \frac{1}{\sqrt{m_p}}$,which implies $\lambda_e > \lambda_p$.

$A$ proton and an electron are accelerated by the same potential difference. Let $\lambda_e$ and $\lambda_p$ denote the de-Broglie wavelengths of the electron and the proton,respectively.

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