Question
A star which is emitting radiation at a wavelength of $$5000\,\mathop {\text{A}}\limits^ \circ $$ is approaching the earth with a velocity of $$1.50 \times {10^6}m/s.$$ The change in wavelength of the radiation as received on the earth is
A.
$$0.25\,\mathop {\text{A}}\limits^ \circ $$
B.
$$2.5\,\mathop {\text{A}}\limits^ \circ $$
C.
$$25\,\mathop {\text{A}}\limits^ \circ $$
D.
$$250\,\mathop {\text{A}}\limits^ \circ $$
Answer :
$$25\,\mathop {\text{A}}\limits^ \circ $$
Solution :
The phenomenon of apparent change in frequency (or wavelength) of the light due to the relative motion between the source of light and the observer is called Doppler effect in light.
$$\eqalign{
& {\text{So,}}\,\,\Delta \lambda = \lambda \times \frac{v}{c}\,......\left( {\text{i}} \right) \cr
& {\text{Given, wavelength}}\,\lambda = 5000\,\mathop {\text{A}}\limits^ \circ \cr
& {\text{Velocity of source}} = 1.5 \times {10^6}\,m/s \cr
& c = 3 \times {10^8}\,m/s \cr
& \therefore \Delta \lambda = 5000 \times \frac{{1.5 \times {{10}^6}}}{{3 \times {{10}^8}}} = 25\,\mathop {\text{A}}\limits^ \circ \cr} $$