When a photon collides with an electron and is deflected, the photons ____

When a photon collides with an electron and is deflected, the photon’s ____ decreases. 1. frequency 2. wavelength

2 months ago

Solution 1

Guest Guest #2391800
2 months ago

Answer:

2. Wavelength

Explanation:

When a photon collides with an electron and is deflected, the photon’s wavelength decreases.

When photon gained energy by the collision of the electron. Its energy and frequency will be increased and its wavelength will be decreased.

Solution 2

Guest Guest #2391801
2 months ago

Answer:

1. Frequency

Explanation:

When a photon collides with an electron, the effect of it is called Compton scattering, where the photons interact with the electron, losing momentum and therefore increase in wavelength and a decrease in frequency.

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Whenever electrons are made to be incident on any anode, they will be interacting with the surface atoms of anode. So these interaction will lead to excitation of electrons of the surface atoms of the anode.

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Find the impulse of a 50. kg object under the following scenarios: a. The object accelerates to 7.5 m/s from rest. b. The object stops from a velocity of 12.0 m/s. C. The object change in velocity from 2.2 m/s to 6.3 d. The object hits the ground with 2.5 m/s and rebounds with the same speed
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  • a. J = 375\, \rm N \cdot s.
  • b. J = -600\; \rm N \cdot s.
  • c. J = 205\; \rm N \cdot s.
  • d. J = -250\; \rm N \cdot s.

Explanation:

When the velocity of an object changes, it would experience an impulse. If the mass of the object stays the same, and that the object moves along a line, the value of the impulse J would be:

J = m \cdot \Delta v, where

  • m is the mass of the object, and
  • \Delta v is the change in the velocity of the object.

On the other hand, the change in the object's velocity can be found with the equation:

\Delta v = v_{\text{final}} - v_\text{initial}.

Note that if m is in \rm kg and \Delta v is in \rm m \cdot s^{-1}, the unit of J would be \rm N \cdot s.

a.

v_\text{initial} = 0\; \rm m \cdot s^{-1}.

v_\text{final} = 7.5\; \rm m \cdot s^{-1}.

\Delta v = v_{\text{final}} - v_\text{initial} = 7.5\; \rm m \cdot s^{-1}.

J = m \cdot \Delta v = 50 \times 7.5 = 375\; \rm N \cdot s.

b.

v_\text{initial} = 12.0\; \rm m \cdot s^{-1}.

v_\text{final} = 0\; \rm m \cdot s^{-1}.

\Delta v = v_{\text{final}} - v_\text{initial} = -12.0\; \rm m \cdot s^{-1}

J = m \cdot \Delta v = 50 \times 12.0 = 600\; \rm N \cdot s.

c.

v_\text{initial} = 2.2\; \rm m \cdot s^{-1}.

v_\text{final} = 6.3\; \rm m \cdot s^{-1}.

\Delta v = v_{\text{final}} - v_\text{initial} = 4.1\; \rm m \cdot s^{-1}.

J = m \cdot \Delta v = 50 \times 4.1 = 201\; \rm N \cdot s.

d.

v_\text{initial} = 2.5\; \rm m \cdot s^{-1}.

v_\text{final} = -2.5\; \rm m \cdot s^{-1}.

Note that v_\text{final} and v_\text{initial} are of opposite signs. The reason is that the object's velocity has changed direction in this period.

\Delta v = v_{\text{final}} - v_\text{initial} = -5.0\; \rm m \cdot s^{-1}.

J = m \cdot \Delta v = 50 \times (-5.0) = -250\; \rm N \cdot s.

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