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2 years ago

Will give correct answer brainliest

Physics

1 answer:

Sav [38]2 years ago

4 0

Answer:

.....

Explanation:

i. sorry? i really hope this helps!

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

Ann [662]

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.

8 0

3 years ago

Why does the density of a substance remain the same for different amount of the substance

tresset_1 [31]

Think of it this way:
-- Any time you have something that means (some number) PER UNIT,
it doesn't matter how many units there are on the table or in the bucket,
because that amount doesn't change the (number) PER UNIT.

-- If oranges cost $1 PER POUND, it doesn't matter how many pounds
you buy, the whole bagful is still $1 PER POUND.

-- If a certain salad dressing has 40 calories PER Tablespoon, it doesn't
matter whether you eat a drop of it or drink the whole jar. You still get
40 calories PER Tablespoon.

-- Density means '(mass) PER unit of volume'. Whether you have a tiny
chip of the substance or a whole truckload of it, there's still the same
amount of mass IN EACH unit of volume.

6 0

3 years ago

What type of weather would be in the middle of the picture, where the barometric pressure is 1000 millibars?

UkoKoshka [18]

It would be D. Hope that helped :)

4 0

3 years ago

Alguien me puede ayudar en mi examen de fisica¿ porfavor es este https://forms.office.com/Pages/ResponsePage.aspx?id=AmBpC6bWZkK

Reptile [31]

Answer:

Chicken Feet Or Butt Whole

Explanation:

Call Me Girls!

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2 years ago

When y⁷ is multiplied by y⁹ the answer is?

Allushta [10]

Answer:

y^16

Explanation:

who need to add the exponents only

7 + 9 = 16

therefore, the answer is y^16

5 0

3 years ago