Magnetic field at the center (at nucleus) of the hydrogen like atom `("atomic number" = z)` due to the motion of electron in nth orbit is proporional to

Options

A. `n^(-3)`
B. `n^(-2)`
C. `n^(-4)`
D. `n^(-5)`

Answer

To find the magnetic field at the center of a hydrogen-like atom due to the motion of an electron in the nth orbit, we can follow these steps:

### Step 1: Understand the Motion of the Electron
The electron in the nth orbit moves in a circular path. This motion can be treated as a circular current loop.

### Step 2: Calculate the Current
The current \( I \) due to the motion of the electron can be defined as the charge passing through a point per unit time. The time \( T \) taken for one complete revolution is given by:
\[ T = \frac{2\pi r_n}{v_n} \]
where \( r_n \) is the radius of the nth orbit and \( v_n \) is the velocity of the electron in that orbit. Therefore, the current \( I \) can be expressed as:
\[ I = \frac{e}{T} = \frac{e}{\frac{2\pi r_n}{v_n}} = \frac{e v_n}{2\pi r_n} \]

### Step 3: Magnetic Field Due to a Current Loop
The magnetic field \( B \) at the center of a circular loop carrying current \( I \) is given by:
\[ B = \frac{\mu_0 I}{2r} \]
where \( \mu_0 \) is the permeability of free space and \( r \) is the radius of the loop.

### Step 4: Substitute the Current into the Magnetic Field Equation
Substituting the expression for current \( I \) into the magnetic field formula, we have:
\[ B = \frac{\mu_0}{2r_n} \cdot \frac{e v_n}{2\pi r_n} = \frac{\mu_0 e v_n}{4\pi r_n^2} \]

### Step 5: Determine the Proportional Relationships
We know from quantum mechanics that:
- The velocity \( v_n \) of the electron is proportional to \( \frac{Z}{n} \) (where \( Z \) is the atomic number).
- The radius \( r_n \) of the nth orbit is proportional to \( \frac{n^2}{Z} \).

Substituting these proportionalities into the equation for \( B \):
- \( v_n \propto \frac{Z}{n} \)
- \( r_n \propto \frac{n^2}{Z} \)

### Step 6: Substitute Proportionalities into the Magnetic Field Expression
Substituting these into the expression for \( B \):
\[ B \propto \frac{\mu_0 e \left(\frac{Z}{n}\right)}{4\pi \left(\frac{n^2}{Z}\right)^2} \]
This simplifies to:
\[ B \propto \frac{\mu_0 e Z^3}{4\pi n^5} \]

### Step 7: Conclusion
Thus, the magnetic field at the center of the hydrogen-like atom due to the motion of the electron in the nth orbit is proportional to:
\[ B \propto \frac{1}{n^5} \]

### Final Answer
The magnetic field at the center of a hydrogen-like atom due to the motion of the electron in the nth orbit is proportional to \( n^{-5} \).

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