A galaxy is moving away from the earth at a speed of `410kms^(-1)`. The shift in the wavelength of a red line at 600 nm is `x xx 10^(-10)m`. The value of x, to the nearest integer, is _____. [Take the value of speed of light c, as `3xx10^(8)ms^(-1)`]

Answer

To solve the problem, we need to find the shift in wavelength (Δλ) of a red line at 600 nm due to the Doppler effect caused by a galaxy moving away from the Earth at a speed of 410 km/s.

### Step-by-Step Solution:

1. **Convert the speed of the galaxy to meters per second**:
\[ V = 410 \, \text{km/s} = 410 \times 10^3 \, \text{m/s} = 4.1 \times 10^5 \, \text{m/s} \]

2. **Identify the speed of light (c)**:
\[ c = 3 \times 10^8 \, \text{m/s} \]

3. **Convert the original wavelength (λ₀) from nanometers to meters**:
\[ \lambda_0 = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} = 6.0 \times 10^{-7} \, \text{m} \]

4. **Use the formula for the shift in wavelength due to the Doppler effect**:
The formula for the shift in wavelength (Δλ) when the source is moving away is given by:
\[ \frac{Δλ}{λ_0} = \frac{V}{c} \]
Rearranging gives:
\[ Δλ = λ_0 \cdot \frac{V}{c} \]

5. **Substituting the values into the equation**:
\[ Δλ = (6.0 \times 10^{-7} \, \text{m}) \cdot \left(\frac{4.1 \times 10^5 \, \text{m/s}}{3 \times 10^8 \, \text{m/s}}\right) \]

6. **Calculate the fraction**:
\[ \frac{4.1 \times 10^5}{3 \times 10^8} = \frac{4.1}{3} \times 10^{-3} \approx 1.3667 \times 10^{-3} \]

7. **Now calculate Δλ**:
\[ Δλ = 6.0 \times 10^{-7} \cdot 1.3667 \times 10^{-3} \approx 8.2002 \times 10^{-10} \, \text{m} \]

8. **Express Δλ in the form of \(x \times 10^{-10} \, \text{m}\)**:
\[ Δλ \approx 8.2 \times 10^{-10} \, \text{m} \]

9. **Identify the value of x**:
\[ x \approx 8.2 \]
Rounding to the nearest integer gives:
\[ x = 8 \]

### Final Answer:
The value of \(x\), to the nearest integer, is **8**.