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? (mp=1.00727u me = 0.00055u)

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43:1

1860:1

414:1

(1860)2:1

answer is A.

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Detailed Solution

λ=h2mqV ⇒λeλp=mpme=1.007270.00055=43

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