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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? ( $m_{p}$ =1.00727u $m_{e}$ = 0.00055u)

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43:1

414:1

1860:1

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

answer is A.

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Detailed Solution

$λ = \frac{h}{\sqrt{2 m q V}} \Rightarrow \frac{λ_{e}}{λ_{p}} = \sqrt{\frac{m_{p}}{m_{e}}} = \sqrt{\frac{1.00727}{0.00055}} = 43$

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