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What is the Be Broglie wavelength of an electron with kinetic energy of 120 eV. |
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Answer» Given, `KE = 120 eV, m = 9.1xx10^(-3)kg, e = 1.6xx10^(-19)c` `lambda = (12.27)/(sqrt(V)) Å = (12.27)/(sqrt(120)) Å = 0.112xx10^(-9) m therefore lambda = 0.112 nm`. |
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