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What are the speed and de broglie wavelength of an electron that has been accelerated by a potent5ial difference of `500 V`? |
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Answer» The KE of the electron under a PD of `500 V `is `1//2 mu^(2) = eV` `u = ((2eV)/(m))^(1//2)` `((2 xx (1.602 xx 10^(-19) C) (500V))/(9.1 xx 10^(-31)kg ))^(1//2)` ` = 1.326 xx 10^(7) m s^(-1)` Using de Broglie equation `lambda = (h)/(mu)` `lambda = (6.26 xx 10^(-34)J s)/((9.1 xx 10^(-31)kg) (1.326 xx 10^(7)ms^(-1))) = 5.5 xx 10^(-11)m` |
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