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3 years ago

• Most of the galaxies in the universe are moving away from

Physics

1 answer:

WARRIOR [948]3 years ago

4 0

The frequency of the light observed from the Earth is $4.945\cdot 10^{14}Hz$

Explanation:

First of all, we start by noticing that the galaxy is receding from Earth (moving away): this means that according to the Doppler effect, the frequency of the light as seen from the Earth must be shorter than the real frequency of the light emitted by the galaxy.

Furthermore, we can quantify the change in frequency of the light using the following equation:

$\frac{\Delta f}{f}=\frac{v}{c}$

where

$\Delta f$ is the change in frequency

f is the real frequency

v is the velocity of recession of the galaxy (negative if the galaxy is moving away)

c is the speed of light

In this problem, we have:

$f=5.00\cdot 10^{14} Hz$

$v=-3325 km/s = -3.325\cdot 10^6 m/s$

$c=3\cdot 10^8 m/s$

Substituting and solving for $\Delta f$ , we find

$\Delta f = \frac{v}{c}f=\frac{-3.325\cdot 10^6}{3\cdot 10^8}(5.00\cdot 10^{14})=-5.542\cdot 10^{12} Hz$

And therefore, the frequency of the light observed from the Earth is

$f'=f+\Delta f = 5.00\cdot 10^{14} +(-5.52\cdot 10^{12})=4.945\cdot 10^{14}Hz$

Learn more about frequency and waves:

brainly.com/question/5354733

brainly.com/question/9077368

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3 years ago

A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl

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Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=157^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.$

Therfore,

$\lambda^{'}-\lambda=4.64 p.m.$

Here, the photon's incident wavelength is $\lamda=7.33pm$

So,

$\lambda^{'}=7.33+4.64=11.97 p.m$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

here, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda$ and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therfore,

$\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.$

This is the de Broglie wavelength of the electron after scattering.

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4 years ago

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3 years ago

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3 years ago

Read 2 more answers

A 5.0-μC charge is placed at the 0 cm mark of a meter stick and a -4.0 μC charge is placed at the 50 cm mark. At what point on a

Maksim231197 [3]

Answer:

The distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

Explanation:

Given that

q₁ = 5 μ C

q₂ = - 4 μ C

The distance between charges = 50 cm

d= 50 cm

Lets take at distance x from the charge μ C ,the electrical field is zero.

That is why the distance from the charge - 4 μ C = 50 - x cm

We know that ,electric field is given as

$E=K\dfrac{q}{r^2}$

$K\dfrac{5\ \mu}{x^2}=K\dfrac{4\mu }{(50-x)^2}\\\\\dfrac{5}{x^2}=\dfrac{4 }{(50-x)^2}\\\\\\5(50-x)^2=4x^2\\(50-x)^2=0.8x^2\\\\50-x =0.89x\\\ x=\dfrac{50}{1.89}\ cm\\\\\\x=26.45\ cm\\$

Therefore the distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

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3 years ago