A spectral line is obtained from a gas discharge tube at 5000Å. If the rms velocity of gas molecules is `10^5 ms^(-1)` , then the width of spectral line will be

Options

A. 3.3Å
B. 4.8Å
C. 7.2Å
D. 9.1Å

Answer

To find the width of the spectral line, we can use the formula that relates the change in wavelength (Δλ) to the velocity of the gas molecules (v) and the speed of light (c). The formula is given by:

\[ \frac{\Delta \lambda}{\lambda} = \frac{v}{c} \]

Where:
- Δλ is the width of the spectral line,
- λ is the wavelength of the spectral line,
- v is the rms velocity of the gas molecules,
- c is the speed of light.

### Step-by-Step Solution:

1. **Identify the given values:**
- Wavelength (λ) = 5000 Å (which is \(5000 \times 10^{-10}\) m for calculations)
- rms velocity (v) = \(10^5\) m/s
- Speed of light (c) = \(3 \times 10^8\) m/s

2. **Convert the wavelength to meters:**
\[ \lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} = 5 \times 10^{-7} \, \text{m} \]

3. **Substitute the values into the formula:**
\[ \frac{\Delta \lambda}{5 \times 10^{-7}} = \frac{10^5}{3 \times 10^8} \]

4. **Calculate the right-hand side:**
\[ \frac{10^5}{3 \times 10^8} = \frac{1}{3} \times 10^{-3} = \frac{1}{3} \times 10^{-3} \, \text{m} \]

5. **Now, solve for Δλ:**
\[ \Delta \lambda = \left(\frac{1}{3} \times 10^{-3}\right) \times (5 \times 10^{-7}) = \frac{5 \times 10^{-10}}{3} \, \text{m} \]

6. **Convert Δλ back to Ångströms:**
\[ \Delta \lambda = \frac{5 \times 10^{-10}}{3} \times 10^{10} \, \text{Å} = \frac{5}{3} \, \text{Å} \approx 1.67 \, \text{Å} \]

### Final Result:
The width of the spectral line (Δλ) is approximately 1.67 Å.