The periodic time of rotation of a certain star is 22 days and its radius is 7 × 10⁸ metres. If the wavelength of light emitted by its surface be 4320 Å, the Doppler shift will be (1 day = 86400 sec)

0.033 Å

$\Delta \lambda=\lambda . \frac{v}{c}$  where  $v = r \omega=r \times\left(\frac{2 \pi}{T}\right)$

∴ $\Delta \lambda=\frac{4320 \times 7 \times 10^8 \times 2 \times 3.14}{3 \times 10^8 \times 22 \times 86400}$ = 0.033 Å