The wavelength of `H_(alpha)` line in the hydrogen spectrum is found to be `6563Å` in the laboratory. If the velocity of the milky way is `1.05xx10^(6)ms^(-1)`, then the wavelength of `H_(alpha)` line in the spectrum of milky way will be

Options

A. `6457Å`
B. `6586Å`
C. `7123Å`
D. `7349Å`

Answer

To solve the problem, we need to determine the wavelength of the H-alpha line in the spectrum of the Milky Way, taking into account the Doppler effect due to the motion of the Milky Way.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Wavelength of H-alpha line in the laboratory, \( \lambda = 6563 \, \text{Å} \)
- Velocity of the Milky Way, \( v = 1.05 \times 10^6 \, \text{m/s} \)
- Speed of light, \( c = 3 \times 10^8 \, \text{m/s} \)

2. **Understand the Doppler Effect**:
The change in wavelength due to the motion of the source (Milky Way) is given by the formula:
\[ \Delta \lambda = \lambda' - \lambda = \frac{v \lambda}{c} \]
where \( \lambda' \) is the observed wavelength, and \( \Delta \lambda \) is the change in wavelength.

3. **Rearranging the Formula**:
We can rearrange the formula to solve for \( \lambda' \):
\[ \lambda' = \lambda + \Delta \lambda = \lambda + \frac{v \lambda}{c} \]
This can be simplified to:
\[ \lambda' = \lambda \left(1 + \frac{v}{c}\right) \]

4. **Substituting the Values**:
Now, substitute the values into the equation:
\[ \lambda' = 6563 \, \text{Å} \left(1 + \frac{1.05 \times 10^6}{3 \times 10^8}\right) \]

5. **Calculating the Fraction**:
First, calculate the fraction:
\[ \frac{1.05 \times 10^6}{3 \times 10^8} = 0.0035 \]

6. **Calculating \( \lambda' \)**:
Now substitute this back into the equation:
\[ \lambda' = 6563 \, \text{Å} \left(1 + 0.0035\right) = 6563 \, \text{Å} \times 1.0035 \]
\[ \lambda' \approx 6585.9 \, \text{Å} \]

7. **Final Result**:
Rounding to the nearest whole number, we find:
\[ \lambda' \approx 6586 \, \text{Å} \]

### Conclusion:
The wavelength of the H-alpha line in the spectrum of the Milky Way is approximately \( 6586 \, \text{Å} \).