A rocket is going away from the earth at a speed of `10^(6)` m/s. If the wavelength of the light wave emited by it be 5700 Å, what will be its Dopper,s shift

Options

A. 200 Å
B. 19 Å
C. 20 Å
D. 0.2 Å

Explanation

To solve the problem of calculating the Doppler shift of light emitted by a rocket moving away from the Earth, we can follow these steps:

### Step 1: Understand the Doppler Shift Formula
The formula for the Doppler shift in wavelength (Δλ) when the source is moving away from the observer is given by:

\[ \Delta \lambda = \frac{v}{c} \cdot \lambda \]

where:
- \( \Delta \lambda \) = change in wavelength (Doppler shift)
- \( v \) = velocity of the source (rocket)
- \( c \) = speed of light
- \( \lambda \) = original wavelength of the emitted light

### Step 2: Identify the Given Values
From the problem, we have:
- \( v = 10^6 \) m/s (speed of the rocket)
- \( \lambda = 5700 \) Å (wavelength of the emitted light)
- \( c = 3 \times 10^8 \) m/s (speed of light)

### Step 3: Convert Wavelength to Meters
Since the speed of light is given in meters per second, we need to convert the wavelength from Angstroms to meters.

1 Å = \( 10^{-10} \) m, thus:

\[ \lambda = 5700 \, \text{Å} = 5700 \times 10^{-10} \, \text{m} = 5.7 \times 10^{-7} \, \text{m} \]

### Step 4: Substitute Values into the Doppler Shift Formula
Now we can substitute the values into the Doppler shift formula:

\[ \Delta \lambda = \frac{10^6 \, \text{m/s}}{3 \times 10^8 \, \text{m/s}} \cdot 5.7 \times 10^{-7} \, \text{m} \]

### Step 5: Calculate the Doppler Shift
Calculating the fraction:

\[ \frac{10^6}{3 \times 10^8} = \frac{1}{300} \approx 0.00333 \]

Now, substituting this value into the equation:

\[ \Delta \lambda = 0.00333 \cdot 5.7 \times 10^{-7} \, \text{m} \]

Calculating this gives:

\[ \Delta \lambda \approx 1.9 \times 10^{-9} \, \text{m} \]

### Step 6: Convert the Result Back to Angstroms
To convert the result back to Angstroms:

\[ \Delta \lambda \approx 1.9 \times 10^{-9} \, \text{m} = 1.9 \times 10^{10} \, \text{Å} = 19 \, \text{Å} \]

### Final Answer
Thus, the Doppler shift is approximately:

\[ \Delta \lambda \approx 19 \, \text{Å} \]

### Conclusion
The correct option is **B) 19 Angstrom**.

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