in a H-atom, the e- is making 6.6*10615 rev/s around the nucleus in an orbit of radius 0.528 Angstrom. the magnetic moment(Am^2) will be 1*10-10. plz explain................

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jyoti dash · Answer

M=IA where M is magnetic field, I is current and A is area so as we know that A= r 2 where r is the radius I=e/t and t=1/f where e is charge t=time and f=frequency substitute the quantity to get the result

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in a H-atom, the e- is making 6.610615 rev/s around the nucleus in an orbit of radius 0.528 Angstrom. the magnetic moment(Am^2) will be 110-10. plz explain................

in a H-atom, the e- is making 6.610615 rev/s around the nucleus in an orbit of radius 0.528 Angstrom. the magnetic moment(Am^2) will be 110-10. plz explain................

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in a H-atom, the e- is making 6.6*10615 rev/s around the nucleus in an orbit of radius 0.528 Angstrom. the magnetic moment(Am^2) will be 1*10-10. plz explain................

in a H-atom, the e- is making 6.6*10615 rev/s around the nucleus in an orbit of radius 0.528 Angstrom. the magnetic moment(Am^2) will be 1*10-10. plz explain................

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in a H-atom, the e- is making 6.610615 rev/s around the nucleus in an orbit of radius 0.528 Angstrom. the magnetic moment(Am^2) will be 110-10. plz explain................

in a H-atom, the e- is making 6.610615 rev/s around the nucleus in an orbit of radius 0.528 Angstrom. the magnetic moment(Am^2) will be 110-10. plz explain................