The apparent wavelength of light from a star moving away from earth is observed to be 0.01% more than its real wavelength. The velocity of star is

Options

A. `120 kms^(-1)`
B. `90 kms^(-1)`
C. `60 kms^(-1)`
D. `30 kms^(-1)`

Answer

To solve the problem, we will use the concept of the Doppler effect for light. The apparent wavelength of light from a star moving away from Earth is observed to be 0.01% more than its real wavelength. We need to find the velocity of the star.

### Step-by-Step Solution:

1. **Understand the given information**:
- The apparent wavelength (\(\lambda'\)) is 0.01% more than the real wavelength (\(\lambda\)).
- This can be expressed as:
\[ \lambda' = \lambda + 0.01\% \text{ of } \lambda = \lambda + 0.0001\lambda = 1.0001\lambda \]

2. **Calculate the change in wavelength (\(\Delta \lambda\))**:
- The change in wavelength (\(\Delta \lambda\)) is given by:
\[ \Delta \lambda = \lambda' - \lambda = 1.0001\lambda - \lambda = 0.0001\lambda \]

3. **Use the Doppler effect formula**:
- The formula relating the change in wavelength to the velocity of the star is:
\[ \frac{\Delta \lambda}{\lambda} = \frac{v}{c} \]
- Where \(v\) is the velocity of the star and \(c\) is the speed of light (\(c = 3 \times 10^8 \text{ m/s}\)).

4. **Substitute \(\Delta \lambda\) into the formula**:
- We know:
\[ \Delta \lambda = 0.0001\lambda \]
- So,
\[ \frac{0.0001\lambda}{\lambda} = \frac{v}{c} \]
- This simplifies to:
\[ 0.0001 = \frac{v}{c} \]

5. **Solve for \(v\)**:
- Rearranging the equation gives:
\[ v = 0.0001 \times c \]
- Substituting the value of \(c\):
\[ v = 0.0001 \times (3 \times 10^8 \text{ m/s}) = 3 \times 10^4 \text{ m/s} \]

6. **Convert \(v\) to kilometers per second**:
- To convert from meters per second to kilometers per second:
\[ v = 3 \times 10^4 \text{ m/s} = 30 \text{ km/s} \]

### Final Answer:
The velocity of the star is \(30 \text{ km/s}\).

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