The speed of an electron in the orbit of hydrogen atom in the ground state is

Options

A. c
B. `(c)/(10)`
C. `(c)/(2)`
D. `(c)/(137)`

Answer

To find the speed of an electron in the orbit of a hydrogen atom in the ground state, we can use the formula derived from the Bohr model of the hydrogen atom. Here’s a step-by-step solution:

### Step 1: Understand the Formula
The speed of an electron in the nth orbit of a hydrogen atom is given by the formula:
\[ V_n = \frac{Z e^2}{2 \epsilon_0 h n} \]
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 ground state, \( n = 1 \)).

### Step 2: Substitute Values
For hydrogen in the ground state:
- \( Z = 1 \)
- \( n = 1 \)

Substituting the values into the formula:
\[ V_1 = \frac{1 \times (1.6 \times 10^{-19})^2}{2 \times (8.85 \times 10^{-12}) \times (6.63 \times 10^{-34}) \times 1} \]

### Step 3: Calculate \( e^2 \)
Calculate \( (1.6 \times 10^{-19})^2 \):
\[ (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \, \text{C}^2 \]

### Step 4: Calculate the Denominator
Calculate the denominator:
\[ 2 \times (8.85 \times 10^{-12}) \times (6.63 \times 10^{-34}) = 1.173 \times 10^{-45} \, \text{C}^2 \text{N m}^2 \text{s} \]

### Step 5: Calculate the Speed
Now substitute back into the formula:
\[ V_1 = \frac{2.56 \times 10^{-38}}{1.173 \times 10^{-45}} \approx 2.18 \times 10^{7} \, \text{m/s} \]

### Step 6: Convert to a Fraction of the Speed of Light
The speed of light \( c \) is approximately \( 3 \times 10^8 \, \text{m/s} \). Now, calculate the ratio \( \frac{V_1}{c} \):
\[ \frac{V_1}{c} = \frac{2.18 \times 10^{7}}{3 \times 10^{8}} \approx 0.0727 \]
This can be expressed as:
\[ V_1 \approx 0.00727 c \approx \frac{1}{137} c \]

### Final Result
Thus, the speed of the electron in the orbit of a hydrogen atom in the ground state is approximately:
\[ V \approx \frac{c}{137} \]