The periodic time of rotation of a certain star is 22 days and its radius is `7 xx 10^(8)` metres . If the wavelength of light emitted by its surface be `4320 Å`, the Doppler shift will be (1 day = 86400 sec )

Options

A. o.36 `overset(o)(A)`
B. 0.03`overset(o)(A)`
C. 3.3 `overset(o)(A)`
D. 33`overset(o)(A)`

Answer

To solve the problem of finding the Doppler shift for the light emitted by a rotating star, we will follow these steps:

### Step 1: Convert the Period of Rotation from Days to Seconds
The period of rotation of the star is given as 22 days. We need to convert this into seconds.

\[ \text{Time period (T)} = 22 \text{ days} \times 86400 \text{ seconds/day} = 22 \times 86400 = 1900800 \text{ seconds} \]

### Step 2: Calculate the Angular Velocity (ω)
The angular velocity (ω) can be calculated using the formula:

\[ \omega = \frac{2\pi}{T} \]

Substituting the value of T:

\[ \omega = \frac{2\pi}{1900800} \approx \frac{6.2832}{1900800} \approx 3.31 \times 10^{-6} \text{ rad/s} \]

### Step 3: Calculate the Linear Velocity (v)
The linear velocity (v) of the star can be calculated using the formula:

\[ v = R \cdot \omega \]

Where R is the radius of the star. Given that \( R = 7 \times 10^8 \) meters:

\[ v = 7 \times 10^8 \text{ m} \times 3.31 \times 10^{-6} \text{ rad/s} \approx 2317 \text{ m/s} \]

### Step 4: Calculate the Doppler Shift (Δλ)
The Doppler shift in wavelength (Δλ) can be calculated using the formula:

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

Where:
- \( \lambda = 4320 \text{ Å} = 4320 \times 10^{-10} \text{ m} \)
- \( c = 3 \times 10^8 \text{ m/s} \)

Substituting the values:

\[ \Delta \lambda = \frac{4320 \times 10^{-10} \text{ m} \cdot 2317 \text{ m/s}}{3 \times 10^8 \text{ m/s}} \]

Calculating the above expression:

\[ \Delta \lambda = \frac{1.000 \times 10^{-6}}{3 \times 10^8} \approx 0.36 \times 10^{-10} \text{ m} = 0.36 \text{ Å} \]

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

\[ \Delta \lambda \approx 0.36 \text{ Å} \]