Answer
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Hint: Atoms which have paired electrons have the magnetic moment zero. Because diamagnetism is the intrinsic property of every material and it is generated due to mutual interaction between the applied magnetic field and orbital motion of electrons.
Complete step by step solution
We know that each electron in an atom is revolving in an orbital around the nucleus. The revolving loop is equivalent to a tiny loop of current. Therefore, it possesses some orbital magnetic moment. \[\overrightarrow{\text{M}}\]
$\overrightarrow{{{\text{M}}_{\text{l}}}}=$current$\times $area of loop
In addition to the orbital motion, every electron is assumed to have a spin motion around its axis. Therefore, magnetic moment is $\overrightarrow{{{\text{M}}_{\text{s}}}}$
The vector sum of $\overrightarrow{{{\text{M}}_{\text{l}}}}$ and$\overrightarrow{{{\text{M}}_{\text{s}}}}$provides the net magnetic moment \[\overrightarrow{\text{M}}\] to the atom.
In diagrammatic material
$\overrightarrow{{{\text{M}}_{\text{l}}}}$and $\overrightarrow{{{\text{M}}_{\text{s}}}}$cancel each other for every atom so that the atom has no net magnetic dipole moment. Therefore, motion of all the electrons in an atom of a diamagnetic material is assured to be reduced to motion of two electrons revolving with same angular velocity in a circular orbit of same radius, but in opposite sense. Hence, the magnetic field of two being equal and opposite, cancel each other (in the absence of any external magnetic field), net magnetic moment is zero.
So the option (D) Equal to zero is the correct answer.
Note
When diamagnetic substances are placed in an external magnetising field, they get feebly magnetised in a direction opposite to the magnetising field. Example: Copper, Lead, Gold, Water, Air, Hydrogen, etc.
Susceptibility of diamagnetic does not change with temperature.
Complete step by step solution
We know that each electron in an atom is revolving in an orbital around the nucleus. The revolving loop is equivalent to a tiny loop of current. Therefore, it possesses some orbital magnetic moment. \[\overrightarrow{\text{M}}\]
$\overrightarrow{{{\text{M}}_{\text{l}}}}=$current$\times $area of loop
In addition to the orbital motion, every electron is assumed to have a spin motion around its axis. Therefore, magnetic moment is $\overrightarrow{{{\text{M}}_{\text{s}}}}$
The vector sum of $\overrightarrow{{{\text{M}}_{\text{l}}}}$ and$\overrightarrow{{{\text{M}}_{\text{s}}}}$provides the net magnetic moment \[\overrightarrow{\text{M}}\] to the atom.
In diagrammatic material
$\overrightarrow{{{\text{M}}_{\text{l}}}}$and $\overrightarrow{{{\text{M}}_{\text{s}}}}$cancel each other for every atom so that the atom has no net magnetic dipole moment. Therefore, motion of all the electrons in an atom of a diamagnetic material is assured to be reduced to motion of two electrons revolving with same angular velocity in a circular orbit of same radius, but in opposite sense. Hence, the magnetic field of two being equal and opposite, cancel each other (in the absence of any external magnetic field), net magnetic moment is zero.
So the option (D) Equal to zero is the correct answer.
Note
When diamagnetic substances are placed in an external magnetising field, they get feebly magnetised in a direction opposite to the magnetising field. Example: Copper, Lead, Gold, Water, Air, Hydrogen, etc.
Susceptibility of diamagnetic does not change with temperature.
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