The orbital magnetic dipole moment (μL) of an electron in an atomic orbital is given by the formula:

μL = -μB * L

where μB is the Bohr magneton and L is the orbital angular momentum quantum number.

For a 3p state, the value of L is 1 (since p corresponds to l=1).

So, the magnitude of the orbital magnetic dipole moment of the electron in the 3p state is:

μL = -μB * 1 = -μB

Therefore, the magnitude of the orbital magnetic dipole moment of the electron in the 3p state is 1 Bohr magneton.

Question

The orbital magnetic dipole moment (μL) of an electron in an atomic orbital is given by the formula:

μL = -μB * L

where μB is the Bohr magneton and L is the orbital angular momentum quantum number.

For a 3p state, the value of L is 1 (since p corresponds to l=1).

So, the magnitude of the orbital magnetic dipole moment of the electron in the 3p state is:

μL = -μB * 1 = -μB

Therefore, the magnitude of the orbital magnetic dipole moment of the electron in the 3p state is 1 Bohr magneton.

Knowee AI · Accepted Answer

The orbital magnetic dipole moment (μL) of an electron in an atomic orbital is given by the formula:

μL = -μB * L

where μB is the Bohr magneton and L is the orbital angular momentum quantum number.

For a 3p state, the value of L is 1 (since p corresponds to l=1).

So, the magnitude of the orbital magnetic dipole moment of the electron in the 3p state is:

μL = -μB * 1 = -μB

Therefore, the magnitude of the orbital magnetic dipole moment of the electron in the 3p state is 1 Bohr magneton.

Find the magnitude of the orbital magnetic dipole moment of the electron in in the 3p state. Provide your answer in terms of the number of Bohr magneton.