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antiseptic1488 [7]

3 years ago

8

Consider Compton Scattering with visible light.A photon with wavelength 500nm scatters backward(theta=180degree) from a free ele

ctron initially at rest.What is the fractional shift in wavelength delta\lamda/\lamda for the photon?

1 answer:

JulijaS [17]3 years ago

5 0

Answer: $4.86(10)^{-12}m$

Explanation:

The Compton Shift $\Delta \lambda$ in wavelength when photons are scattered is given by the following equation:

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda'=500 nm=500(10)^{-9} m$ is the wavelength of the scattered photon

$\lambda_{o}$ is the wavelength of the incident photon

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}.c}$ , being $h=4.136(10)^{-15}eV.s$ the Planck constant, $m_{e}$ the mass of the electron and $c=3(10)^{8}m/s$ the speed of light in vacuum.

$\theta=180\°$ the angle between incident phhoton and the scatered photon.

$\Delta \lambda=2.43(10)^{-12} m (1-cos(180\°))$ (2)

$\Delta \lambda=4.86(10)^{-12}m$ (3) This is the shift in wavelength

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3 years ago

Verdich [7]

5)

In physics, forces are interactions that are able to change the velocity of an object.

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6)

In physics, arrows are used to represent vector quantities. Therefore, they are also used to represent forces.

In particular, when a vector quantity is represented by an arrowr:

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7)

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We can rewrite the Newton in terms of fundamental units only. We can do it starting from the equation:

$F=ma$

where

F is the force

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a is the acceleration

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Therefore, 1 N corresponds to:

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B)

Gravity is an attractive force that exists between all objects that have mass. See more explanations about gravity in part 4).

3)

Mass is a scalar quantity; it gives us a measure of the "amount of matter" contained in an object.

The SI unit of the mass is the kilogram (kg).

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Moreover, the mass is an intrinsec property of an object: therefore, it does not depend on the location of the object. So, an object has always the same mass, either it is on Earth or on another planet.

On the other hand, the force of gravity on an object depends on its location, so it changes.

4)

As we said in part 3), gravity is an attractive force that exists between all objects that have mass.

The magnitude of the force of gravity between two objects is given by the Universal Law of gravitation:

$F=\frac{Gm_1 m_2}{r^2}$

where

G is the gravitational constant

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7 0

4 years ago

Somebody please explain how to solve this. Thanks in advance!

Jobisdone [24]

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3 years ago

The viewing screen in a double-slit experiment with monochromatic light. Fringe C is the central maximum. The fringe separation

makvit [3.9K]

Answer:

<em>Part A</em><em>:</em>

a) If the wavelength of the light is decreased the fringe spacing Δy will decrease.

<em>Part B</em><em>:</em>

b) If the spacing between the slits is decreased the fringe spacing Δy will increase.

<em>Part C</em><em>:</em>

a) If the distance to the screen is decreased the fringe spacing will decrease.

<em>Part D</em><em>:</em>

The dot in the center of fringe E is $920\ x\ 10^{-9} m$ farther from the left slit than from the right slit.

Explanation:

In the double-slit experiment there is a clear contrast between the dark and bright fringes, that indicate destructive and constructive interference respectively, in the central peak and then is less so at either side.

The position of bright fringes in the screen where the pattern is formed can be calculated with

$\vartriangle y =\frac{m \lambda L}{d}$

$m=0,\pm 1,\pm 2,\pm 3,.....$

m is the order number.
$\lambda$ is the wavelength of the monochromatic light.
L is the distance between the screen and the two slits.
d is the distance between the slits.

Part A: a) In the above equation for the position of bright fringes we can see that if the wavelength of the light $\lambda$ is decreased the overall effect will be that the fringes are going to be closer. That means that the fringe spacing Δy will decrease.

Part B: b) In the above equation for the position of bright fringes we can see that if the spacing between the slits d is decreased the fringes are going to be wider apart. That means the fringe spacing Δy will increase.
Part C: a) In the above equation we can see that if the distance to the screen L is decreased the fringes are going to be closer. That means the fringe spacing Δy will decrease.

Part D: We are told that the central maximum is the fringe C that corresponds with m=0. That means that fringe E corresponds with the order number m=2 if we consider it to be the second maximum at the rigth of the central one. To calculate how much farther from the left slit than from the right slit is a dot located at the center of the fringe E in the screen we use the condition for constructive interference. That says that the path length difference Δr between rays coming from the left and right slit must be $\vartriangle r=m \lambda$

We simply replace the values in that equation :

$\vartriangle r= m \lambda =2.\ 460\ nm$

$\vartriangle r= 920\ x\ 10^{-9} m$

The dot in the center of fringe E is $920\ x\ 10^{-9}m$ farther from the left slit than from the right slit.

3 0

3 years ago