Astronomers have observed that light coming from far away starts do not show the same spectrum as observed on Earth. Each wavelength is shifted slightly towards the red end of the spectrum. It is observed that the `1^(st)` line of Lyman series shows 121 nm on Earth, but from the star it shows 122nm. The speed of the star with respect to Earth is close to

Options

A. `2.5xx10^(6)ms^(-1)`
B. `2xx10^(6)ms^(-1)`
C. `1.5xx10^(6)ms^(-1)`
D. `0.5xx10^(6)ms^(-1)`

Answer

To solve the problem, we will use the concept of redshift, which is the phenomenon where light from an object moving away from an observer is shifted to longer wavelengths (towards the red end of the spectrum).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Wavelength observed on Earth (λ₀) = 121 nm
- Wavelength observed from the star (λ) = 122 nm

2. **Calculate the Change in Wavelength (Δλ):**
\[ Δλ = λ - λ₀ = 122 \, \text{nm} - 121 \, \text{nm} = 1 \, \text{nm} \]

3. **Use the Redshift Formula:**
The formula relating the change in wavelength to the speed of the star is:
\[ \frac{Δλ}{λ₀} = \frac{V}{C} \]
where:
- \( V \) = speed of the star relative to Earth
- \( C \) = speed of light (approximately \( 3 \times 10^8 \, \text{m/s} \))

4. **Substitute the Values into the Formula:**
\[ \frac{1 \, \text{nm}}{121 \, \text{nm}} = \frac{V}{3 \times 10^8 \, \text{m/s}} \]

5. **Solve for V:**
Rearranging the equation gives:
\[ V = C \cdot \frac{Δλ}{λ₀} \]
Substituting the known values:
\[ V = 3 \times 10^8 \, \text{m/s} \cdot \frac{1 \, \text{nm}}{121 \, \text{nm}} \]

6. **Convert nm to m for Calculation:**
Since \( 1 \, \text{nm} = 10^{-9} \, \text{m} \):
\[ V = 3 \times 10^8 \, \text{m/s} \cdot \frac{1 \times 10^{-9} \, \text{m}}{121 \times 10^{-9} \, \text{m}} = 3 \times 10^8 \cdot \frac{1}{121} \]

7. **Calculate the Final Speed:**
\[ V \approx 2.48 \times 10^6 \, \text{m/s} \]

### Final Answer:
The speed of the star with respect to Earth is approximately \( 2.5 \times 10^6 \, \text{m/s} \).

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