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An electron, an alpha particle, and a proton have the same de-Broglie wavelength. If their kinetic energies are E1, E2 and E3 respectively, which relation is correct?

1. E1 is greater than E3, which is greater than E2
2. E2 is greater than E3, which is greater than E1
3. E1 is greater than E2, which is greater than E3
4. E2 is equal to E3

Correct answer: E1 is greater than E3, which is greater than E2

Solution

Equal de-Broglie wavelength means equal momentum p (lambda = h/p). Kinetic energy E = p^2/(2m) is inversely proportional to mass. Since m(electron) < m(proton) < m(alpha), E1 > E3 > E2.

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