An electron of mass m_e and a proton of mass
Question:

An electron of mass $m_{e}$ and a proton of mass $m_{p}=1836 \mathrm{~m}_{\mathrm{e}}$ are moving with the same speed.

The ratio of their de Broglie wavelength $\frac{\lambda_{\text {eectron }}}{\lambda_{\text {proton }}}$

will be :

  1. 1836

  2. 1

  3. 918

  4. $\frac{1}{1836}$


Correct Option: 1

Solution:

$\frac{\lambda_{e}}{\lambda_{p}}=\frac{\frac{\mathrm{h}}{\mathrm{m}_{e} \mathrm{v}}}{\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{p}} \mathrm{v}}}=1836$

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