Discussion
An electron and a proton are accelerated through the same potential. The ratio of their de Broglie wavelengths, \(\lambda_{e}/\lambda_{p}\) is :
Answer: A
The ratio is (mp/me)1/2.
A particle of charge q accelerated through a potential V gains kinetic energy qV. The electron and the proton carry the same magnitude of charge, so they gain the same kinetic energy. Its momentum and de Broglie wavelength are
\[ \begin{aligned} p &= \sqrt{2mqV} \\ \lambda &= \frac{h}{p} = \frac{h}{\sqrt{2mqV}} \end{aligned} \]
With h, q and V the same for both, λ is proportional to 1/√m:
\[ \begin{aligned} \frac{\lambda_{e}}{\lambda_{p}} &= \sqrt{\frac{m_{p}}{m_{e}}} \\ &\approx \sqrt{1836} \approx 43 \end{aligned} \]
me/mp is the inverse of the mass ratio and gives the wrong direction (the lighter electron has the longer wavelength). mp/me without the square root holds for equal speeds, not equal energies, and (mp/me)2 has no basis.
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