The de-Broglie wavelength associated with an electron and a proton were calculated by

Question:
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 $?\left(\mathrm{~m}_{\mathrm{P}}=1.00727 \mathrm{u}, \mathrm{m}_{\mathrm{e}}=0.00055 \mathrm{u}\right)$

$1860: 1$
$(1860)^{2}: 1$
$41.4: 1$
$43: 1$

Correct Option: , 4

Solution:

$\lambda=\frac{h}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}$

$\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}}}$

$\frac{\lambda_{\mathrm{e}}}{\lambda_{\mathrm{P}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{P}}}{\mathrm{m}_{\mathrm{e}}}}=\sqrt{1831.4}=42.79$

The de-Broglie wavelength associated with an electron and a proton were calculated by

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