In Bohr's orbit of hydrogen atom m kg is more of an electron and e couomb is the charge on it. The ratio ( in SI units) of magnetic dipole moment to that of the angular momentum of electron is :

Options

A. e/2m
B. e/m
C. 2e/m
D. 2e/3m

Explanation

To solve the problem of finding the ratio of the magnetic dipole moment (μ) to the angular momentum (L) of an electron in a Bohr orbit of a hydrogen atom, we can follow these steps:

### Step 1: Understand the Magnetic Dipole Moment
The magnetic dipole moment (μ) of a current loop can be expressed as:
\[ \mu = I \cdot A \]
where \(I\) is the current and \(A\) is the area of the loop.

### Step 2: Determine the Current
The current \(I\) due to the electron moving in a circular orbit can be defined as the charge passing through a point per unit time. For an electron with charge \(e\) moving in a circular orbit with period \(T\):
\[ I = \frac{e}{T} \]

### Step 3: Calculate the Area of the Orbit
The area \(A\) of the circular orbit with radius \(r\) is given by:
\[ A = \pi r^2 \]

### Step 4: Substitute into the Magnetic Dipole Moment Formula
Now substituting \(I\) and \(A\) into the formula for μ:
\[ \mu = I \cdot A = \left(\frac{e}{T}\right) \cdot (\pi r^2) = \frac{e \pi r^2}{T} \]

### Step 5: Relate Period to Angular Momentum
The period \(T\) can be related to the angular momentum \(L\). The angular momentum \(L\) of the electron is given by:
\[ L = mvr \]
where \(v\) is the velocity of the electron. The period \(T\) can also be expressed in terms of the velocity and radius:
\[ T = \frac{2\pi r}{v} \]

### Step 6: Substitute for T in μ
Substituting \(T\) back into the equation for μ gives:
\[ \mu = \frac{e \pi r^2 v}{2\pi r} = \frac{e r v}{2} \]

### Step 7: Expressing v in terms of L
From the angular momentum expression, we can express \(v\) as:
\[ v = \frac{L}{mr} \]
Substituting this into the equation for μ:
\[ \mu = \frac{e r}{2} \cdot \frac{L}{mr} = \frac{eL}{2m} \]

### Step 8: Find the Ratio of μ to L
Now, we can find the ratio of the magnetic dipole moment to the angular momentum:
\[ \frac{\mu}{L} = \frac{\frac{eL}{2m}}{L} = \frac{e}{2m} \]

### Conclusion
Thus, the ratio of the magnetic dipole moment to the angular momentum of the electron in a Bohr orbit of a hydrogen atom is:
\[ \frac{\mu}{L} = \frac{e}{2m} \]