Question

A star, which is emitting radiation at a wavelength of $5000\,\mathring{A}$ is approaching the earth with a velocity of $1.5 \times 10^5\, m/s.$ The change in wavelength of the radiation as received on the earth is

• A. $25\,\mathring{A}$
• B. $100 \,\mathring{A}$
• C. zero
• D. $2.5\,\mathring{A}$

Answer 1

Answer: (A)

$\lambda_{0}=\sqrt{\frac{1+\beta}{1-\beta}} \lambda_{s}$
$\lambda_{0}$ is observed wave length. $\lambda$ is emitted wave length and $\beta=\frac{ v }{ c }$
where $c$ is speed of light and $v =1.5 \times 10^{6} m / s$
$\delta \lambda=\lambda_{0}-\lambda_{ s }=\lambda_{ C }\left(1-\sqrt{\frac{1+\beta}{1-\beta}}\right) $
$=\lambda\left(1-\sqrt{\frac{1+1.5 \times 10^{6} / 3 \times 10^{8}}{1-1.5 \times 10^{6} / 3 \times 10^{8}}}\right)=25\,\mathring{A}$