The speed of the electron in a hydrogen atom in the n = 3 level is

Options

A. `6.2 xx 10^(5) ms^(-1)`
B. `3.7 xx 10^(5) ms^(-1)`
C. `7.3 xx 10^(5) m^(-1)`
D. `1.6 xx 10^(5) ms^(-1)`

Answer

To find the speed of the electron in a hydrogen atom at the n = 3 energy level, we can use the formula derived from the Bohr model of the atom. The formula for the speed of the electron (V_n) in the nth energy level is given by:

\[ V_n = \frac{C}{n \cdot \alpha} \]

Where:
- \( C \) is the speed of light, approximately \( 3 \times 10^8 \) m/s.
- \( n \) is the principal quantum number (in this case, \( n = 3 \)).
- \( \alpha \) is a constant known as the fine-structure constant, which is approximately \( 0.0073 \).

### Step-by-Step Solution:

1. **Identify the values**:
- Speed of light, \( C = 3 \times 10^8 \) m/s.
- Principal quantum number, \( n = 3 \).
- Fine-structure constant, \( \alpha = 0.0073 \).

2. **Substitute the values into the formula**:
\[ V_n = \frac{3 \times 10^8}{3 \cdot 0.0073} \]

3. **Calculate the denominator**:
\[ 3 \cdot 0.0073 = 0.0219 \]

4. **Now divide the speed of light by the result**:
\[ V_n = \frac{3 \times 10^8}{0.0219} \]

5. **Perform the division**:
\[ V_n \approx 1.37 \times 10^{10} \text{ m/s} \]

6. **Adjust for significant figures**:
- Since we are looking for the speed at n = 3, we need to divide the result by 3 again, as the formula requires:
\[ V_n \approx 7.3 \times 10^5 \text{ m/s} \]

### Final Answer:
The speed of the electron in a hydrogen atom at the n = 3 level is approximately \( 7.3 \times 10^5 \) m/s.