The speed of the electron in the 1st orbit of the hydrogen atom in the ground state is (c is the veloicty of light)

Options

A. `(c)/(1.37)`
B. `(c)/(1370)`
C. `(c)/(13.7)`
D. `(c)/(137)`

Explanation

To find the speed of the electron in the first orbit of the hydrogen atom in the ground state, we can use the Bohr model of the hydrogen atom. According to this model, the speed of the electron in the nth orbit can be calculated using the following formula:

\[ v_n = \frac{Z e^2}{2 \epsilon_0 h} \cdot \frac{1}{n} \]

Where:
- \( v_n \) is the speed of the electron in the nth orbit,
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( e \) is the charge of the electron (\( 1.6 \times 10^{-19} \) C),
- \( \epsilon_0 \) is the permittivity of free space (\( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)),
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \)),
- \( n \) is the principal quantum number (for the ground state of hydrogen, \( n = 1 \)).

### Step-by-Step Solution:

1. **Identify Constants**:
- For hydrogen, \( Z = 1 \).
- Use the known constants:
- \( e = 1.6 \times 10^{-19} \, \text{C} \)
- \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)
- \( h = 6.63 \times 10^{-34} \, \text{Js} \)

2. **Substitute Values into the Formula**:
\[ v_1 = \frac{(1)(1.6 \times 10^{-19})^2}{2(8.85 \times 10^{-12})(6.63 \times 10^{-34})} \]

3. **Calculate the Numerator**:
\[ \text{Numerator} = (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \, \text{C}^2 \]

4. **Calculate the Denominator**:
\[ \text{Denominator} = 2 \times (8.85 \times 10^{-12}) \times (6.63 \times 10^{-34}) = 1.174 \times 10^{-45} \, \text{C}^2 \cdot \text{Js} \]

5. **Calculate the Speed**:
\[ v_1 = \frac{2.56 \times 10^{-38}}{1.174 \times 10^{-45}} \approx 2.18 \times 10^{7} \, \text{m/s} \]

6. **Express Speed in Terms of Speed of Light**:
- The speed of light \( c \) is approximately \( 3 \times 10^8 \, \text{m/s} \).
\[ \text{Fraction of } c = \frac{v_1}{c} = \frac{2.18 \times 10^{7}}{3 \times 10^8} \approx 0.0727 \, c \]

### Final Answer:
The speed of the electron in the first orbit of the hydrogen atom in the ground state is approximately \( 2.18 \times 10^{7} \, \text{m/s} \) or about \( 0.0727 \, c \).