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An electron beam energes from an accelerator with kinetic energy 100eV. What is its de-Broglie wavelength? `[m = 9.1 xx 10^(-31)kg, h = 6.6 xx 10^(-34)Js, 1eV = 1.6 xx 10^(-19)J]` |
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Answer» Kinetic energy of electron `= 100eV` `= 100 xx 1.6 xx 10^(-19) J = 1.6 xx 10^(-17)J` we know, `lambda = (h)/(sqrt(2KE.m))` `= (6.626 xx 10^(-34))/(sqrt(2xx 1.6 xx 10^(-17) xx 9.1 xx 10^(-31))) = 1.228 xx 10^(-10)m` |
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