JEE MAIN - Physics (2021 - 25th February Evening Shift - No. 4)
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 :
1
1836
$${1 \over {1836}}$$
918
Explanation
Given mass of electron = me
Mass of proton = mp
$$ \therefore $$ given mp = 1836 me
From de-Broglie wavelength
$$\lambda = {h \over p} = {h \over {mv}}$$
$${{{\lambda _e}} \over {{\lambda _p}}} = {{{m_p}} \over {{m_e}}}$$
$$ = {{1836{m_e}} \over {{m_e}}}$$
$${{{\lambda _e}} \over {{\lambda _p}}} = 1836$$
Mass of proton = mp
$$ \therefore $$ given mp = 1836 me
From de-Broglie wavelength
$$\lambda = {h \over p} = {h \over {mv}}$$
$${{{\lambda _e}} \over {{\lambda _p}}} = {{{m_p}} \over {{m_e}}}$$
$$ = {{1836{m_e}} \over {{m_e}}}$$
$${{{\lambda _e}} \over {{\lambda _p}}} = 1836$$
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