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2 years ago

Types of polarization physics

Physics

1 answer:

Sidana [21]2 years ago

8 0

Answer:

Linear polarization.

Circular polarization.

Elliptical polarization.

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X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to

lys-0071 [83]

Answer:

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

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

b) $\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

c) $E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

Explanation

Part a

For this case we can use the Compton shift equation given by:

$\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

Part b

For this cas we can calculate the wavelength of the phton with this formula:

$\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

$E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

3 0

3 years ago

Read 2 more answers

A ball dropped from a window strikes the ground 2.76 seconds later. How high is the window above the ground

sweet [91]

Answer:

37.33m

Explanation:

Using the equation of motion

S = ut + 1/2gt^2

Time t = 2.76secs

g = 9.8m/s^2

S = 0 + 1/2(9.8)(2.76)^2

S = 4.9*7.6176

S = 37.33

Hence the window is 37.33m above the ground

7 0

3 years ago

Rachel hits a golf ball into the air. What type of motion is the ball's path?

sladkih [1.3K]

Answer:

omg

the ground is BREAKING

Explanation:

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3 years ago

What are the two applications of electron beam welding

Georgia [21]

First of all, there are not <u>just</u> two applications that are solely applicable to the electron beam welding process. There are MANY.

Please visit out website at the URL below and you can click the "View Application" button under each listed Industry segment to view case studies of commonly EB welded applications.

https://www.ptreb.com/electron-beam-welding-applications

And for more general information on our welding process, we have an informational section you can peruse as well:

https://www.ptreb.com/electron-beam-welding-information

Good luck with your assignment- we are glad to hear they are teaching about EBW in high school!!!

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3 years ago

Please help fill in the blank Earth’s axis passes through the (blank) and south poles.

LuckyWell [14K]

Answer:

North

Explanation:

Have a nice day :)

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3 years ago