The de-Broglie wavelength associated with an electron and a proton were calculated by accelerating them through same potential of 100 V. What should nearly be the ratio of their wavelengths? (\(m_p = 1.00727\ \mathrm{u}\), \(m_e = 0.00055\ \mathrm{u}\))
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- \(41.4 : 1\)
- \((1860)^2 : 1\)
- \(1860 : 1\)
- \(43 : 1\)
The Correct Option is D
Solution and Explanation
\(\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}} = \sqrt{\frac{1.00727}{0.00055}} = \sqrt{1831.4} = 42.8 \approx 43\).
So the ratio is about \(43 : 1\).
Answer: (4) \(43 : 1\)
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