The de-Broglie wavelength associated with an electron and a proton were
calculated by accelerating them through same potential of $100 \mathrm{~V}$. What
should nearly be the ratio of their wavelengths?
$\left(\mathrm{m}_{\mathrm{P}}=1.00727 \mathrm{u}, \mathrm{m}_{\mathrm{e}}=0.00055 \mathrm{u}\right)$
Correct Option: , 4
(4)
$\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}$
$\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{m_{2}}{m_{1}}}$
$\frac{\lambda_{\mathrm{e}}}{\lambda_{\mathrm{p}}}=\sqrt{\frac{\mathrm{mP}}{\mathrm{m}_{\mathrm{e}}}}=\sqrt{1831.4}=42.79$
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