Chapter 33, Problem 46GP

Solution

a.

The speed of galaxy.

  $v=1.49\times {10}^8$ m/sec

Given:

A galaxy is moving away from earth.

The blue hydrogen line wavelength is 434nm when emitted from the galaxy butit is measured on 650nm on Earth.

Formula used:

  $\nabla \lambda =\frac{v\lambda }{c}$

Here,

  $\nabla \lambda$ is the change in wavelength.

v is velocity.

  $\lambda$ is the original wavelength.

c is the speed of light.

Calculation:

From $\nabla \lambda =\frac{v\lambda }{c}$

  $v=\frac{c\nabla \lambda }{\lambda }$

  $\nabla \lambda =650-434=216\text{nm}$

  $\lambda =434\text{nm}$

  $c=3\times {10}^8\text{m}/\text{sec}$

Substituting the values,

  $v=\frac{3\times {10}^8\times 216}{434}$

  $v=\frac{648\times {10}^8}{434}$

  $v=1.49\times {10}^8\text{m}/\text{sec}$

b.

The distance of galaxy from earth.

  $d=6.25\times {10}^{25}\text{m}$

Given:

A galaxy is moving away from earth.

The blue hydrogen line wavelength is 434nm when emitted from the galaxy butit is measured on 650nm on Earth.

Formula used:

  $v=H_{0}d$

Here,

H₀ is the Hubble constant.

d is the distance.

v is the speed.

Calculation:

From $v=H_{0}d$

  $d=\frac{v}{H_0}$

  $v=1.49\times {10}^8$ m/sec

Value of H₀ is 73.8km/sec/Mpc

Number of km in Mpc is $3.09\times {10}^{19}$

For converting Hubble’s constant in the unit of per second , we divide the value of hubble’s constant (km/sec/Mpc )by $3.09\times {10}^{19}$

So Hubble’s constant in sec-1 is $\frac{73.8}{3.09\times {10}^{19}}=23.88\times {10}^{-19}$

  $d=\frac{1.49\times {10}^8}{23.88\times {10}^{-19}}$

  $d=0.0625\times {10}^{27}\text{m}$

  $d=6.25\times {10}^{25}\text{m}$