In hydrogen spectrum the wavelength of `H_(a)` line is `656nm, ` where in the spectrum of a distance galaxy `H_(a)` line wavelength is `706 nm` . Estimated speed of the galaxy with respect to earth is ,

Options

A. `2xx10^(8)m//s`
B. `2xx10^(7)m//s`
C. `2xx10^(6)m//s`
D. `2xx10^(5)m//s`

Answer

To solve the problem of estimating the speed of a distant galaxy with respect to Earth based on the observed shift in the wavelength of the Hα line, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Wavelength of Hα line on Earth (λ) = 656 nm
- Wavelength of Hα line in the galaxy (λ') = 706 nm

2. **Calculate the Wavelength Shift (Δλ):**
\[ \Delta \lambda = \lambda' - \lambda = 706 \, \text{nm} - 656 \, \text{nm} = 50 \, \text{nm} \]

3. **Convert Wavelengths to Meters:**
- Since 1 nm = \(10^{-9}\) m, we convert the wavelengths:
\[ \lambda = 656 \, \text{nm} = 656 \times 10^{-9} \, \text{m} \]
\[ \Delta \lambda = 50 \, \text{nm} = 50 \times 10^{-9} \, \text{m} \]

4. **Use the Formula for Velocity (V):**
The formula relating the speed of the galaxy (V) to the wavelength shift is:
\[ \Delta \lambda = \frac{V}{C} \cdot \lambda \]
Rearranging gives:
\[ V = \frac{C \cdot \Delta \lambda}{\lambda} \]
Where:
- \(C\) = speed of light = \(3 \times 10^8 \, \text{m/s}\)

5. **Substitute the Values:**
\[ V = \frac{3 \times 10^8 \, \text{m/s} \cdot (50 \times 10^{-9} \, \text{m})}{656 \times 10^{-9} \, \text{m}} \]

6. **Calculate the Velocity:**
\[ V = \frac{3 \times 10^8 \cdot 50}{656} \, \text{m/s} \]
\[ V \approx \frac{15000000000}{656} \, \text{m/s} \approx 22830.5 \, \text{m/s} \]

7. **Convert to Scientific Notation:**
\[ V \approx 2.28 \times 10^4 \, \text{m/s} \approx 2 \times 10^7 \, \text{m/s} \]

8. **Conclusion:**
The estimated speed of the galaxy with respect to Earth is approximately \(2 \times 10^7 \, \text{m/s}\).

### Final Answer:
The estimated speed of the galaxy with respect to Earth is \(2 \times 10^7 \, \text{m/s}\).

---