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vampirchik [111]
3 years ago
8

The “Compton effect” showed

Physics
2 answers:
Blizzard [7]3 years ago
8 0
The answer is the last choice, the particle nature of light. The compton effect showed that a charged electron scatters the photon results in decreased energy but increase in wavelength.
Yanka [14]3 years ago
3 0

Answer:

the particle nature of light

Explanation:

As we know that by the formula of Compton Effect we have

\lambda' - \lambda = \frac{h}{mc}(1 - cos\theta)

here in this formula we will say that when light of some given wavelength \lambda incident on a particle then after deviating from the particle the light will change its direction by some angle \theta.

And the wavelength of the particle will also changed to some different value as \lambda'

so here as per particle theory we will use the theory of momentum conservation as per which light of given momentum when hit a particle then it will transfer it momentum to the particle and deviated by some lesser momentum.

Due to decrease in momentum the change in wavelength of light is given by above formula

so this verify the particle nature of light

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If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle
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This question is incomplete, the complete question is;

The electric force due to a uniform external electric field causes a torque of magnitude 20.0 × 10⁻⁹ N⋅m on an electric dipole oriented at 30° from the direction of the external field. The dipole moment of the dipole is 7.5 × 10⁻¹² C⋅m.

What is the magnitude of the external electric field?

If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle?

Answer:

- the magnitude of the external electric field is 5333.3 N/C

- the magnitude of the charge on each particle is 3.0 × 10⁻¹² C  ≈ 3 nC

Explanation:

Given that;

Torque = 20.0 × 10⁻⁹ N⋅m

dipole moment = 7.5 × 10⁻¹²

∅ = 30°

The moment T of restoring couple is;

T = PEsin∅

E = T/Psin∅

we substitute

E = 20.0 × 10⁻⁹ N⋅m / (7.5 × 10⁻¹²) sin(30°)

E = 20.0 × 10⁻⁹ / 3.75 × 10⁻¹²

E =  5333.3 N/C

Therefore, the magnitude of the external electric field is 5333.3 N/C

The dipole moment is given by the expression;

p = ql

q = p / l

given that l = 2.5 mm = 0.0025 m

we substitute

q = 7.5 × 10⁻¹² / 0.0025

q = 3.0 × 10⁻¹² C ≈ 3 nC

Therefore, the magnitude of the charge on each particle is 3.0 × 10⁻¹² C ≈ 3 nC

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2 years ago
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