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The electron in a hydrogen atom moves in a circular orbit of radius 5 xx 10_(-11) m with a speed of 0.67 xx 10_(8) m//s Then |
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Answer» the frequency of revolutions of the electron is `6 xx 10^(-15) rev//s` `i = FE, where f = frequency =(omega)/(2pi)=(v)/(2pir)=(0.6xx10^(6)xxpi)/(2pixx5xx10^(-11))=6xx10^(15)rev//s(0.6xx10^(6)xxpi)/(2pixx5xx10^(-11))=6xx10^(15)rev//s` `i=6xx10^(15)xx1.6xx10^(-19)`= `0.96 xx 10^(-3)A = 0.96 mA` therefore Hence, all the options are correct |
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