All categories

3 years ago

As the scattering angle of the photon increases, what happens to the wavelength associated with the photon?

Physics

1 answer:

frez [133]3 years ago

6 0

As the scattering angle of the photon increases, the wavelength associated with the photon increases.

<h3><u>Explanation:</u></h3>

The particle with quantum mechanical property is known as Compton wavelength. The wavelength of a photon increases during collision. When the scattering angle of the photon is 0 degree then the photon's wavelength increases by 0 and when the scattering angle is 180 degree then the wavelength of the photon will become double. This is known as Compton wavelength.

When a photon undergoes collision process, the photo loses its energy and this energy is transferred to the electrons. This causes energy of the photon to decrease and thus the frequency also decreases. Thus, the wavelength of the photon will increase.

You might be interested in

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 58.8m/s2 . The accel

Zanzabum

Split the operation in two parts. Part A) constant acceleration 58.8m/s^2, Part B) free fall.

Part A)
Height reached, y = a*[t^2] / 2 = 58.8 m/s^2 * [7.00 s]^2 / 2 = 1440.6 m

Now you need the final speed to use it as initial speed of the next part.

Vf = Vo + at = 0 + 58.8m/s^2 * 7.00 s = 411.6 m/s

Part B) Free fall

Maximum height, y max ==> Vf = 0

Vf = Vo - gt ==> t = [Vo - Vf]/g = 411.6 m/s / 9.8 m/s^2 = 42 s

ymax = yo + Vo*t - g[t^2] / 2

ymax = 1440.6 m + 411.6m/s * 42 s - 9.8m/s^2 * [42s]^2 /2
ymax = 1440.6 m + 17287.2m - 8643.6m = 10084.2 m

Answer: ymax = 10084.2m

8 0

3 years ago

The first law of thermodynamics relates the heat transfer into or out of a system to the change of internal and the work done on

Artyom0805 [142]

Answer:

The heat transferred into the system is 183.5 J.

Explanation:

The first law of thermodynamics relates the heat transfer into or out of a system to the change of internal and the work done on the system, through the following equations.

ΔU = Q - W

where;

ΔU is the change in internal energy

Q is the heat transfer into the system

W is the work done by the system

Given;

ΔU = 155 J

W = 28.5 J

Q = ?

155 = Q - 28.5

Q = 155 + 28.5

Q = 183.5 J

Therefore, the heat transferred into the system is 183.5 J.

4 0

2 years ago

A certain wave motion has a frequency of 10 cycles / s and a wavelength of 3 m. so its propagation speed is

kodGreya [7K]

3.5m is ur answer ask for more questions anytime

5 0

3 years ago

In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these s

riadik2000 [5.3K]

Answer:

Part a)

$\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}$

Part b)

$\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}$

Explanation:

As we know that the see saw bar is massless so here torque due to two masses is given as

$\tau = I\alpha$

here we will have

$\tau = (m_1g - m_2g)(\frac{L}{2})$

now we will have inertia of two masses given as

$I = (m_1 + m_2)(\frac{L}{2})^2$

now we have

$I = (m_1 + m_2)\frac{L^2}{4}$

now the angular acceleration is given as

$\alpha = \frac{\tau}{I}$

so we have

$\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}$

Part b)

Now if the rod is not massles then we will have total inertia given as

$I = (m_1 + m_2)(\frac{L}{2})^2 + \frac{m_{bar}L^2}{12}$

so we will have

$I = (m_1 + m_2)\frac{L^2}{4} + \frac{m_{bar}L^2}{12}$

now the acceleration is given as

$\alpha = \frac{\tau}{I}$

$\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}$

7 0

3 years ago

A force acting over a large area will exert less pressure per square inch than the same force acting over a smaller area.

babunello [35]

Answer:

True

Explanation:

Pressure is defined as:

$p=\frac{F}{A}$

where

F is the magnitude of the force perpendicular to the surface

A is the surface

Therefore, pressure is inversely proportional to the area of the surface:

$p\propto \frac{1}{A}$

this means that, assuming that the forces in the two situations (which have same magnitude) are both applied perpendicular to the surface, the force exerted over the smaller area will exert a greater pressure. Hence, the statement"

<em>"A force acting over a large area will exert less pressure per square inch than the same force acting over a smaller area"</em>

is true.

8 0

3 years ago