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 follow these steps:

### Step 1: Identify the formula for the speed of an electron in an orbit
The speed \( v \) of an electron in any orbit is given by the formula:
\[ v = 2.12 \times 10^6 \frac{Z}{n} \text{ m/s} \]
where:
- \( Z \) is the atomic number,
- \( n \) is the principal quantum number.

### Step 2: Substitute the values for hydrogen atom
For a hydrogen atom:
- The atomic number \( Z = 1 \) (since hydrogen has one proton),
- In the ground state, the principal quantum number \( n = 1 \).

Substituting these values into the formula:
\[ v = 2.12 \times 10^6 \frac{1}{1} \text{ m/s} = 2.12 \times 10^6 \text{ m/s} \]

### Step 3: Convert the speed to a fraction of the speed of light
The speed of light \( c \) is approximately \( 3 \times 10^8 \) m/s. To express the speed of the electron as a fraction of the speed of light, we can calculate:
\[ \frac{v}{c} = \frac{2.12 \times 10^6}{3 \times 10^8} \]

### Step 4: Simplify the expression
Calculating this gives:
\[ \frac{v}{c} = \frac{2.12}{3} \times 10^{-2} \approx 0.00007067 \]
This can be approximated as:
\[ \frac{v}{c} \approx \frac{1}{137} \]

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