Question

23. A star, which is emitting radiation at a wavelength
of 5000 A, is approaching the earth with a velocity
of 1.5 * 105 m/s. The change in wavelength of the
radiation as received on the earth is
(1) 25 A
(2) 100 Å
(3) Zero
(4) 2.5 Å .

Answer 1

Answer:

The change in wavelength is 25 A° .

Explanation:

Given as :

The wavelength of radiation emitted by star = $\lambda _1$ = 5000 A°

The velocity with which it approaching = v = 1.5 × $10^{5}$ m/s

Let The change in wavelength after receiving on earth = Δ$\lambda$

The wavelength of radiation on earth = $\lambda _2$

According to question

$\lambda _2$ = $\lambda _1$ × [ $\dfrac{c - v}{c}$ ]      , where c is speed of light = 3 × $10^{8}$

Or, $\lambda _2$ = $\lambda _1$ × [ 1 - $\dfrac{v}{c}$ ]

Or, $\dfrac{\lambda _2}{\lambda _1}$ = [ 1 - $\dfrac{v}{c}$ ]

Or, $\dfrac{v}{c}$ = 1 -  $\dfrac{\lambda _2}{\lambda _1}$

Or,  $\dfrac{v}{c}$ = $\dfrac{\lambda _2 - \lambda _1}{\lambda _1}$

Or, $\dfrac{v}{c}$ = $\dfrac{\Delta \lambda }{\lambda _1}$

i.e $\Delta \lambda$ = $\lambda _1$ × $\dfrac{v}{c}$

or, $\Delta \lambda$ = 5000 A° × $\dfrac{1.5\times 10^{6}}{3\times 10^{8}}$

or, $\Delta \lambda$ = 5000 A° × 0.005

∴  $\Delta \lambda$ = 25 A°

So, The change in wavelength =  $\Delta \lambda$ = 25 A°

Hence, The change in wavelength is 25 A° . Answer