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

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

  1. (1) $1860: 1$

  2. (2) $(1860)^{2}: 1$

  3. (3) $41.4: 1$

  4. (4) $43: 1$


Correct Option: , 4

Solution:

(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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