The ratio of the speed of the electrons in the ground state of hydrogen to the speed of light in vacuum is

Options

A. `1//2`
B. `2/137`
C. `1//137`
D. `1//237`

Explanation

To find the ratio of the speed of the electrons in the ground state of hydrogen to the speed of light in vacuum, we can follow these steps:

### Step 1: Identify the formula for the speed of electrons in the ground state of hydrogen.
The speed of an electron in the nth orbit of a hydrogen atom is given by the formula:
\[ v_n = \frac{2.19 \times 10^6 \, \text{m/s}}{n} \]
For the ground state, \(n = 1\).

### Step 2: Calculate the speed of the electron in the ground state.
Substituting \(n = 1\) into the formula, we get:
\[ v_1 = \frac{2.19 \times 10^6 \, \text{m/s}}{1} = 2.19 \times 10^6 \, \text{m/s} \]

### Step 3: Identify the speed of light in vacuum.
The speed of light in vacuum, denoted as \(c\), is:
\[ c = 3 \times 10^8 \, \text{m/s} \]

### Step 4: Calculate the ratio of the speed of the electron to the speed of light.
We need to find the ratio \( \frac{v_1}{c} \):
\[ \frac{v_1}{c} = \frac{2.19 \times 10^6 \, \text{m/s}}{3 \times 10^8 \, \text{m/s}} \]

### Step 5: Simplify the ratio.
Calculating the ratio:
\[ \frac{v_1}{c} = \frac{2.19}{3} \times \frac{10^6}{10^8} = \frac{2.19}{3} \times 10^{-2} \]
Calculating \( \frac{2.19}{3} \):
\[ \frac{2.19}{3} \approx 0.73 \]
Thus:
\[ \frac{v_1}{c} \approx 0.73 \times 10^{-2} = 7.3 \times 10^{-3} \]

### Step 6: Compare with the options provided.
The options given are:
1. \( \frac{1}{2} \)
2. \( \frac{2}{137} \)
3. \( \frac{1}{137} \)
4. \( \frac{1}{237} \)

Calculating \( \frac{1}{137} \):
\[ \frac{1}{137} \approx 0.0073 = 7.3 \times 10^{-3} \]
This matches our calculated ratio.

### Final Answer:
The ratio of the speed of the electrons in the ground state of hydrogen to the speed of light in vacuum is:
\[ \frac{1}{137} \]

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