Velocity of star is `10^(6)m//s` and frequency of emitted light is `4.5X10^(14)Hz`. If star is moving away, then apparent frequency will be

Options

A. `4.5Hz`
B. `4.5xx10^(16)Hz`
C. `4.485X10^(14)Hz`
D. `4.5X10^(8)Hz`

Explanation

To find the apparent frequency of light emitted by a star that is moving away from us, we can use the formula for the Doppler effect for light. The formula is given by:

\[ \nu' = \nu \left(1 - \frac{v}{c}\right) \]

Where:
- \(\nu'\) = apparent frequency
- \(\nu\) = emitted frequency
- \(v\) = velocity of the star (moving away)
- \(c\) = speed of light

### Step-by-Step Solution:

1. **Identify the given values:**
- Velocity of the star, \(v = 10^6 \, \text{m/s}\)
- Frequency of emitted light, \(\nu = 4.5 \times 10^{14} \, \text{Hz}\)
- Speed of light, \(c = 3 \times 10^8 \, \text{m/s}\)

2. **Substitute the values into the Doppler effect formula:**
\[ \nu' = 4.5 \times 10^{14} \left(1 - \frac{10^6}{3 \times 10^8}\right) \]

3. **Calculate the fraction \(\frac{v}{c}\):**
\[ \frac{v}{c} = \frac{10^6}{3 \times 10^8} = \frac{1}{300} \approx 0.00333 \]

4. **Substitute this value back into the formula:**
\[ \nu' = 4.5 \times 10^{14} \left(1 - 0.00333\right) \]
\[ \nu' = 4.5 \times 10^{14} \times 0.99667 \]

5. **Calculate the apparent frequency:**
\[ \nu' \approx 4.5 \times 10^{14} \times 0.99667 \approx 4.485 \times 10^{14} \, \text{Hz} \]

### Final Answer:
The apparent frequency of the light emitted by the star moving away from us is approximately:
\[ \nu' \approx 4.485 \times 10^{14} \, \text{Hz} \]