"An electron of mass me and a proton of mass mp = 1836 me are moving with the same speed. The ratio of their de Broglie wavelength ${{{}^\lambda electron} \over {{}^\lambda proton}}$ will be :"

Question

Accepted Answer

"Given mass of electron = m_e

Mass of proton = m_p

$\therefore$ given m_p = 1836 m_e

From de-Broglie wavelength

$\lambda = {h \over p} = {h \over {mv}}$

${{{\lambda _e}} \over {{\lambda _p}}} = {{{m_p}} \over {{m_e}}}$

$= {{1836{m_e}} \over {{m_e}}}$

${{{\lambda _e}} \over {{\lambda _p}}} = 1836$"

An electron of mass m_e and a proton of mass m_p = 1836 m_e are moving with the same speed. The ratio of their de Broglie wavelength ${{{}^\lambda electron} \over {{}^\lambda proton}}$ will be :

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An electron of mass me and a proton of mass mp = 1836 me are moving with the same speed. The ratio of their de Broglie wavelength ${{{}^\lambda electron} \over {{}^\lambda proton}}$ will be :

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An electron of mass m_e and a proton of mass m_p = 1836 m_e are moving with the same speed. The ratio of their de Broglie wavelength ${{{}^\lambda electron} \over {{}^\lambda proton}}$ will be :