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In the hydrogen atom, an electron revolve around a proton in a circular orbit of radiu 0.53 A. The radial acceleration and the angular velocity of the electron are......and…………. (m_(e) = 9.1 xx 10^(-31) kg, e = 1.6 xx 10^(-19)C) |
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Answer» Solution :Suppose, charge on a proton and an electron is e The electric force between them at DISTANCE d will be: `F = (ke^(2))/r^(2)` `=(9 xx 10^(9) xx (1.6 xx 10^(-19))^(2))/(0.53 xx 10^(-10))^(2)` `= 82.02 xx 10^(-9)` `therefore F = 8.2 xx 10^(-8)` N The centripetal ACCELERATION `=("Centripetal force")/("mass")` `therefore a_( c) = F/m` [Mass of electron, `m_(e) = 9.1 xx 10^(-31) kg`] `=(8.1 xx 10^(-8))/(9.1 xx 10^(-31))` `therefore a_( c) = 0.9 xx 10^(23) m//s^(2)` Centripetal force `=(MV^(2))/r = (mr^(2)omega^(2))/r [therefore v= romega]` `therefore F = mromega^(2)` `therefore omega = [F/(mr)]^(1/2)` `=[17 xx 10^(32)]^(1//2)` `therefore omega = 4.1 xx 10^(16)` rad/s |
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