A distant star is moving constantly towards the Earth at a speed of `2.3 xx 10^4 ms^(-1)`. A shift of `0.5 Å` in wavelengths is observed from the Earth. Calculate the actual value of wavelength emitted from the star if the Earth is considered at rest with respect to the distant star.

Answer

To solve the problem, we will use the Doppler effect formula for light, which relates the change in wavelength (\( \Delta \lambda \)) to the speed of the source (\( v \)) and the speed of light (\( c \)).

### Step-by-Step Solution:

1. **Identify the given values:**
- Speed of the star (\( v \)) = \( 2.3 \times 10^4 \, \text{m/s} \)
- Shift in wavelength (\( \Delta \lambda \)) = \( 0.5 \, \text{Å} = 0.5 \times 10^{-10} \, \text{m} \)
- Speed of light (\( c \)) = \( 3 \times 10^8 \, \text{m/s} \)

2. **Use the Doppler effect formula:**
The formula relating the change in wavelength to the speed of the source and the speed of light is given by:
\[ \frac{\Delta \lambda}{\lambda} = \frac{v}{c} \]
Rearranging this gives:
\[ \lambda = \frac{c \cdot \Delta \lambda}{v} \]

3. **Substitute the known values into the equation:**
\[ \lambda = \frac{(3 \times 10^8 \, \text{m/s}) \cdot (0.5 \times 10^{-10} \, \text{m})}{2.3 \times 10^4 \, \text{m/s}} \]

4. **Calculate the wavelength:**
- First, calculate the numerator:
\[ 3 \times 10^8 \times 0.5 \times 10^{-10} = 1.5 \times 10^{-2} \, \text{m} \]
- Now, divide by the speed of the star:
\[ \lambda = \frac{1.5 \times 10^{-2} \, \text{m}}{2.3 \times 10^4 \, \text{m/s}} \approx 6.52174 \times 10^{-7} \, \text{m} \]

5. **Convert the wavelength to angstroms:**
Since \( 1 \, \text{Å} = 10^{-10} \, \text{m} \):
\[ \lambda \approx 6.52174 \times 10^{-7} \, \text{m} = 6521.74 \, \text{Å} \]

6. **Round the answer:**
The actual wavelength emitted from the star is approximately:
\[ \lambda \approx 6500 \, \text{Å} \]

### Final Answer:
The actual value of the wavelength emitted from the star is approximately \( 6500 \, \text{Å} \).

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