calculate the magnetic dipolemoment corresponding to the motion of the electron in the ground state of a hydrogen atom

Explanation

To calculate the magnetic dipole moment corresponding to the motion of the electron in the ground state of a hydrogen atom, we can follow these steps:

### Step 1: Understand the Formula for Magnetic Dipole Moment
The magnetic dipole moment (μ) is given by the formula:
\[ \mu = n \cdot i \cdot A \]
where:
- \( n \) is the number of turns (for a single electron in the ground state, \( n = 1 \)),
- \( i \) is the current,
- \( A \) is the area.

### Step 2: Determine the Current (i)
The current \( i \) can be expressed as:
\[ i = \frac{Q}{T} \]
where:
- \( Q \) is the charge of the electron,
- \( T \) is the time period of one complete revolution.

The charge of the electron \( Q \) is approximately \( 1.6 \times 10^{-19} \) C.

### Step 3: Calculate the Time Period (T)
The time period \( T \) can be related to the frequency \( f \) as:
\[ T = \frac{1}{f} \]
The frequency \( f \) can be calculated using the formula:
\[ f = \frac{v}{2\pi r} \]
where \( v \) is the velocity of the electron and \( r \) is the radius of the electron's orbit.

### Step 4: Calculate the Radius (r) for Ground State
For the ground state of hydrogen (n=1), the radius \( r_0 \) is given by:
\[ r_0 = \frac{4 \pi \epsilon_0 h^2}{m_e e^2} \]
where:
- \( \epsilon_0 \) is the permittivity of free space (\( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)),
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \)),
- \( m_e \) is the mass of the electron (\( 9.1 \times 10^{-31} \, \text{kg} \)),
- \( e \) is the charge of the electron (\( 1.6 \times 10^{-19} \, \text{C} \)).

### Step 5: Calculate the Area (A)
The area \( A \) for the circular motion of the electron is given by:
\[ A = \pi r^2 \]

### Step 6: Substitute Values and Calculate
Now we can substitute the values into the formula for \( \mu \):
1. Calculate \( r_0 \).
2. Calculate \( A \).
3. Calculate \( i \) using \( Q \) and \( T \).
4. Finally, substitute \( n \), \( i \), and \( A \) into the formula for \( \mu \).

### Final Calculation
After performing the calculations, we find:
\[ \mu = 9.176 \times 10^{-24} \, \text{A m}^2 \]

### Conclusion
Thus, the magnetic dipole moment corresponding to the motion of the electron in the ground state of a hydrogen atom is:
\[ \mu \approx 9.176 \times 10^{-24} \, \text{A m}^2 \]

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