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The ratio (in S.I. units) of magnetic dipole moment to that of the angular of electron of mass m kg and charge e coulomb in Bohr's orbit of hydrogen atom is : |
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Answer» `(c)/(2m)` `=(e(pi r^(2)))/(2pi) omega=er^(2) omega=(er^(2) omega)/(2) ........(1)` Angular momentum `=mvr =mr^(2) omega` `("magnetic dipole moment")/("Angular momentum")=(er^(2)omega)/(2) XX (1)/(mr^(2) omega)=(e)/(2m)` |
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