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Calculate the deBroglie wavelength of an electron travilling at `1%` of the speed of the light |
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Answer» The mass of electron is `9.11 xx 10^(-31) kg ` One percent of the speed of light is `s = (1.100)(3.00 xx 10^(8) m s^(-1))` `= 3.00 xx 10^(6) m s^(-1)` The momentum of the electron is given by `p = mv = (9.11 xx 10^(-31) kg) (3.00 xx 10^(6) m s^(-1))` ` = 2.733 xx 10^(-24) kg m s^(-1)` The de Broglie wavelength of this electron is `lambda = (h)/(p) = (6.626 xx 10^(-34) Js)/(2.733 xx 10^(-24) kg m s^(-1))` `= 2.424 xx 10^(-10) m = 242 .4 "pm" ` The wavelength is of atomic dimenssion |
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