Find the maximum angular speed of the electron of a hydrogen atoms in a statoonary orbit

Explanation

To find the maximum angular speed of the electron in a hydrogen atom in a stationary orbit, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship**: We know from Bohr's model that the angular momentum of the electron in a stationary orbit is quantized. The formula for angular momentum (L) is given by:
\[ L = mvr = \frac{nh}{2\pi} \]
where \(m\) is the mass of the electron, \(v\) is its linear velocity, \(r\) is the radius of the orbit, \(n\) is the principal quantum number, and \(h\) is Planck's constant.

2. **Relate linear velocity to angular velocity**: The linear velocity \(v\) can be expressed in terms of angular velocity \(\omega\) as:
\[ v = r\omega \]
Substituting this into the angular momentum equation gives:
\[ m(r\omega)r = \frac{nh}{2\pi} \]
This simplifies to:
\[ mr^2\omega = \frac{nh}{2\pi} \]

3. **Solve for angular velocity**: Rearranging the equation to solve for \(\omega\):
\[ \omega = \frac{nh}{2\pi mr^2} \]

4. **Substitute known values**: For the hydrogen atom in its ground state, \(n = 1\). The values we need are:
- \(h = 6.63 \times 10^{-34} \, \text{Js}\)
- \(m = 9.1 \times 10^{-31} \, \text{kg}\)
- The Bohr radius \(r = 0.53 \, \text{Å} = 0.53 \times 10^{-10} \, \text{m}\)

5. **Calculate \(r^2\)**:
\[ r^2 = (0.53 \times 10^{-10})^2 = 0.2809 \times 10^{-20} \, \text{m}^2 \]

6. **Plug in the values**:
\[ \omega = \frac{1 \cdot 6.63 \times 10^{-34}}{2\pi \cdot 9.1 \times 10^{-31} \cdot 0.2809 \times 10^{-20}} \]

7. **Calculate the denominator**:
\[ 2\pi \cdot 9.1 \times 10^{-31} \cdot 0.2809 \times 10^{-20} \approx 5.094 \times 10^{-50} \]

8. **Final calculation for \(\omega\)**:
\[ \omega \approx \frac{6.63 \times 10^{-34}}{5.094 \times 10^{-50}} \approx 1.30 \times 10^{16} \, \text{rad/s} \]

9. **Final result**:
\[ \omega \approx 4.13 \times 10^{16} \, \text{rad/s} \]

### Final Answer:
The maximum angular speed of the electron in a hydrogen atom in a stationary orbit is approximately \(4.13 \times 10^{16} \, \text{rad/s}\).