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An electron moves around the nucleus in a hydrogen atom of radius 0.51 Å , with a velocity of 2 xx 10^(5) m s^(-1) . Calculate the following : (i) the equivalent current due to orbital motion of electron, (ii) the megnetic field produced at the centre of the nucleus, (iii) the magnetic associated with the electron. |
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Answer» Solution :It is given that `r = 0.51 Å = 5.1 xx 10^(-11) m, v = 2 xx 10^(5) ms^(-1) , and q = e = 1.6 xx 10^(-19)C` `l = (ev)/(2 pi r) = (1.6 xx 10^(-19) xx 2 xx 10^(5))/(2 xx 3.14 xx 5.1 xx 10^(-11)) = 10^(-4)A` (ii) Magnetic field at the centre of the nucleus `B = (mu_0 I)/(2r) = (4 pi xx 10^(-7) xx 10^(-4))/(2 xx 5.1 xx 10^(-11)) = 1.25 T`. (iii) The magnetic moment ASSOCIATED with the electron `m = IA = I pir^2 = 10^(-4) xx 3.14 xx (5.1 xx 10^(-11))^2 = 8.2 xx 10^(-25) A m^(2)`. |
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