Question

The $6563\mathring{A}$ line emitted by hydrogen atom in a star is found to be red shifted by $5\mathring{A}$. the speed with which the star is receding from the earth is:

• A. $17.3\times {10}^{3}m/s$
• B. $4.29\times {10}^{7}m/s$
• C. $3.39\times {10}^{5}m/s$
• D. $2.29\times {10}^{5}m/s$

Answer 1

Answer: (D)

$2.29\times {10}^{5}m/s$

$\frac{△v}{v}=\frac{V_{radiant}}{C}$
$v=\frac{C}{\lambda }$
$d.v=\frac{C}{{\lambda }^2}.dv=\frac{\frac{x}{{\lambda }^2}}{\frac{C}{\lambda }}=\frac{d\lambda }{\lambda }$

$\frac{d\lambda }{\lambda }=\frac{V_{radial}}{C}$

$\frac{5\times {10}^{-10}}{6563\times {10}^{-10}}=\frac{V_{radial}}{3\times {10}^8}$

Calculate $V_{radial}=\frac{5}{6563}\times 3\times {10}^8$

$V_{radial}=\frac{15}{6563}\times {10}^8$

$V_{radial}=\frac{15}{6.56}\times {10}^5$

$V_{radial}=2.29\times {10}^{5}m/s$