The ratio of magnetic dipole moment of an electron of charge e and mass m in the Bohr orbit in hydrogen to the angular momentum of the electron in the orbit is:

Options

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

Answer

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

### Step 1: Define the magnetic dipole moment (μ)
The magnetic dipole moment (μ) of an electron moving in a circular orbit can be expressed as:
\[ \mu = I \cdot A \]
where \(I\) is the current and \(A\) is the area of the circular orbit.

### Step 2: Calculate the current (I)
The current \(I\) due to the electron moving in a circular path can be defined as the charge passing through a point per unit time. For an electron of charge \(e\) moving with velocity \(v\) in a circular orbit of radius \(r\):
\[ I = \frac{e}{T} \]
where \(T\) is the time period of one complete revolution. The time period \(T\) can be expressed as:
\[ T = \frac{2\pi r}{v} \]
Thus, substituting \(T\) in the expression for current:
\[ I = \frac{e \cdot v}{2\pi r} \]

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

### Step 4: Substitute into the magnetic dipole moment equation
Now substituting \(I\) and \(A\) into the equation for magnetic dipole moment:
\[ \mu = I \cdot A = \left(\frac{e \cdot v}{2\pi r}\right) \cdot \pi r^2 = \frac{e \cdot v \cdot r}{2} \]

### Step 5: Define the angular momentum (L)
The angular momentum \(L\) of the electron in the orbit is given by:
\[ L = m \cdot v \cdot r \]

### Step 6: Find the ratio of magnetic dipole moment to angular momentum
Now we can find the ratio of magnetic dipole moment to angular momentum:
\[ \frac{\mu}{L} = \frac{\frac{e \cdot v \cdot r}{2}}{m \cdot v \cdot r} \]
Here, \(v\) and \(r\) cancel out:
\[ \frac{\mu}{L} = \frac{e}{2m} \]

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

### Final Answer
The ratio of the magnetic dipole moment of an electron of charge \(e\) and mass \(m\) in the Bohr orbit in hydrogen to the angular momentum of the electron in the orbit is \(\frac{e}{2m}\).