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The electron in the hydrogen atom is moving with a speed of `2.5 xx 10^(6)` m/s in an orbit of radius 0.5 `Å`. Magnetic moment of the revolving electron is |
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Answer» Given : `v=2.3xx10^(6)m//s, r=0.53 "Å"=0.53 xx10^(-10)m` Time period of revolution is given by, `T=(2pi r)/(v)` Substituting the value, we get `T=(2xx3.14 xx0.53xx10^(-10))/(2.3xx10^(6))` `T=(1.06xx3.142xx10^(-16))/(2.3)` `=(3.33052xx10^(-16))/(2.3)` `T=1.448xx10^(-16)s` |
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