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Question

(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

$\lambda_e > \lambda_p$

Vedclass · Answer

(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 elementary charge.Since $h$,$e$,and $V$ are constant for both particles,we have $\lambda \propto \frac{1}{\sqrt{m}}$.We know that the mass of an electron $(m_e)$ is much smaller 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 $\lambda_e > \lambda_p$.Thus,the correct option is $(C)$.

$A$ proton and an electron are accelerated by the same potential difference. If their de-Broglie wavelengths are $\lambda_p$ and $\lambda_e$ respectively,then:

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