A particle is travelling \(4\) times as fast as an electron. Assuming the ratio of de-Broglie wavelength of a particle to that of electron is \(2: 1,\) the mass of the particle is

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Correct Answer

(D) \(\frac{1}{8}\) times the mass of \(e ^{-}\)

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\(\lambda=\frac{ h }{ p }\)

\(\frac{\lambda_{ p }}{\lambda_{ e }}=\frac{ p _{ e }}{ p _{ p }}=\frac{ m _{ e } v _{ e }}{ m _{ p } v _{ p }}\)

\(2=\frac{ m _{ e }}{ m _{ p }}\left(\frac{ v _{ e }}{4 v _{ e }}\right)\)

\(\therefore m _{ p }=\frac{ m _{ e }}{8}\)

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