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

Options

A. `6.2xx10^(5)rad//s`
B. `4.1xx10^(16) rad//s`
C. `2.4xx10^(10) rad//s`
D. `9.2xx10^(10) rad//s`

Answer

To find the maximum angular speed (ω) of the electron in a stationary orbit of a hydrogen atom, we can use the formula for angular momentum and the relationship between linear velocity and angular velocity.

### Step-by-Step Solution:

1. **Understanding Angular Momentum**:
The angular momentum (L) of an electron in a stationary orbit is given by:
\[ L = mvr \]
where:
- \( m \) = mass of the electron
- \( v \) = linear velocity of the electron
- \( r \) = radius of the orbit

2. **Bohr's Quantization Condition**:
According to Bohr's model, the angular momentum is quantized and given by:
\[ L = n \frac{h}{2\pi} \]
where:
- \( n \) = principal quantum number (for the ground state of hydrogen, \( n = 1 \))
- \( h \) = Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \))

3. **Equating the Two Expressions for Angular Momentum**:
Setting the two expressions for angular momentum equal gives:
\[ mvr = n \frac{h}{2\pi} \]

4. **Relating Linear Velocity to Angular Velocity**:
The linear velocity \( v \) can be expressed in terms of angular velocity \( \omega \):
\[ v = r\omega \]
Substituting this into the angular momentum equation gives:
\[ mr^2\omega = n \frac{h}{2\pi} \]

5. **Solving for Angular Speed (ω)**:
Rearranging the equation to solve for \( \omega \):
\[ \omega = \frac{n h}{2\pi m r^2} \]

6. **Substituting Known Values**:
- For hydrogen atom, \( n = 1 \)
- \( h = 6.626 \times 10^{-34} \, \text{Js} \)
- Mass of the electron \( m = 9.1 \times 10^{-31} \, \text{kg} \)
- Bohr radius \( r = 0.53 \times 10^{-10} \, \text{m} \)

Substituting these values into the equation:
\[ \omega = \frac{1 \times 6.626 \times 10^{-34}}{2\pi \times 9.1 \times 10^{-31} \times (0.53 \times 10^{-10})^2} \]

7. **Calculating ω**:
- First, calculate \( (0.53 \times 10^{-10})^2 \):
\[ (0.53 \times 10^{-10})^2 = 0.2809 \times 10^{-20} = 2.809 \times 10^{-21} \]
- Now substitute this back into the equation for \( \omega \):
\[ \omega = \frac{6.626 \times 10^{-34}}{2\pi \times 9.1 \times 10^{-31} \times 2.809 \times 10^{-21}} \]
- Calculate the denominator:
\[ 2\pi \times 9.1 \times 10^{-31} \times 2.809 \times 10^{-21} \approx 5.09 \times 10^{-51} \]
- Finally, calculate \( \omega \):
\[ \omega \approx \frac{6.626 \times 10^{-34}}{5.09 \times 10^{-51}} \approx 1.30 \times 10^{17} \, \text{rad/s} \]

8. **Final Answer**:
The maximum angular speed of the electron in a stationary orbit of a hydrogen atom is approximately:
\[ \omega \approx 4.1 \times 10^{16} \, \text{rad/s} \]