Question

A hydrogen line in a star's spectrum has a frequency of 6.17*10^14 Hz when stationary. In Rigel's Spectrum, it is shifted downward by 4.26*10^10Hz. What is Rigel's velocity relative to us? m/s

Answer 1

Answer:

The correct answer will be "2.97 × 10⁸ m/s".

Explanation:

The give values are:

Observed frequency,  $F=4.26\times 10^{10} \ Hz$

Original frequency,  $F_{0}=6.17\times 10^{14} \ Hz$

Let the velocity of Rigel be = V m/s

As we know,

⇒   $F_{0}=F\sqrt{\frac{C+V}{C-V}}$

On putting the values, we get

⇒   $\frac{6.17\times 10^{14} }{4.26\times 10^{10} } =\sqrt{\frac{C+V}{C-V} }$

⇒   $\frac{C+V}{C-V}=(14483.56808)^2$

⇒   $\frac{C+V}{C-V}=209773744.2$

⇒   $C+V=209773744.2 \ C-209773744.2 \ V$

⇒   $C-209773744.2 \ C=209773744.2 \ V-V$

⇒   $V=\frac{209773743.2}{209773745.2} \ C$

⇒   $V=0.99 \ C$

So that the Rigel's velocity will be "0.99 C".

Now,

⇒  $V=0.99\times 3\times 10^8$

⇒      $=2.97\times 10^8 \ m/s$