The electron moves in a circular orbit around the nucleus, creating a loop current.
The current \( I \) is:
\[ I = \frac{\text{Charge per revolution}}{\text{Time for one revolution}} \]
The charge of an electron is \( e \) and time period \( T \) is:
\[ T = \frac{2\pi r}{v} \]
Thus,
\[ I = \frac{e}{T} = \frac{e}{\frac{2\pi r}{v}} = \frac{ev}{2\pi r} \]
The magnetic moment \( \mu \) is given by:
\[ \mu = I \times A \]
Since the electron follows a circular path, the area is:
\[ A = \pi r^2 \]
Thus,
\[ \mu = \frac{ev}{2\pi r} \times \pi r^2 \]
\[ \mu = \frac{evr}{2} \]
The angular momentum of the electron is:
\[ L = mvr \]
where \( m \) is the mass of the electron.
Since:
\[ \mu = \frac{evr}{2} \]
Replacing \( vr \) using \( L \):
\[ \mu = \frac{e}{2m} L \]
Thus, the magnetic moment associated with the electron is:
\[ \mu = \frac{e}{2m} L \]
(b) If \( \vec{L} \) is the angular momentum of the electron, show that:
\[ \vec{\mu} = -\frac{e}{2m} \vec{L} \]
The sum of the spin-only magnetic moment values (in B.M.) of $[\text{Mn}(\text{Br})_6]^{3-}$ and $[\text{Mn}(\text{CN})_6]^{3-}$ is ____.
A racing track is built around an elliptical ground whose equation is given by \[ 9x^2 + 16y^2 = 144 \] The width of the track is \(3\) m as shown. Based on the given information answer the following: 
(i) Express \(y\) as a function of \(x\) from the given equation of ellipse.
(ii) Integrate the function obtained in (i) with respect to \(x\).
(iii)(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration.
OR
(iii)(b) Write the coordinates of the points \(P\) and \(Q\) where the outer edge of the track cuts \(x\)-axis and \(y\)-axis in first quadrant and find the area of triangle formed by points \(P,O,Q\).