Quantum optics

1. What is a photon?

Solution:

heories on the nature of light:

a.) Light is a mechanical wave of the ether. (Huygens, Young)
b.) Light is a stream of particles (corpuscles) emitted from the source. (Newton)
c.) Light is an electromagnetic wave with a frequency of about 3.8·10¹⁴ Hz to 7.8·10¹⁴ Hz, which propagates continuously from the source. (Maxwell)
d.) The energy of radiation (light) is not distributed continuously in space, but consists of a certain number of discrete packets of quanta, which can only be emitted and absorbed as whole units. A quantum of radiation (light) is called a photon. (Planck, Einstein)

Each photon has:
Energy: $E = h \cdot f = m \cdot c^2 = p \cdot c \quad h = 6.625 \cdot 10^{-34} \, J \cdot s \quad c = 3.10^8 \, m \cdot s^{-1}$
Momentum $p = m \cdot c = h / \lambda$

Wave–particle duality:
de Broglie: Every moving particle with mass $m$ and velocity $v$ has also a wave with wavelength

\lambda = \frac{h}{m \cdot v}

Compton: Each photon can be considered a massive particle with zero rest mass, which exists only at speed $c$ .

Compton experiment:

E = h \cdot (f_1 - f_2)

\Delta \lambda = \frac{h}{m_e \cdot c} (1 - \cos \vartheta) \quad \vartheta = \text{angle of photon deflection}

Photoelectric effect (photoemission) is the release of electrons from metals by incident radiation (photons):

internal (electrons are emitted from atoms but remain within the metal)
external (electrons are emitted from the surface of the metal and move in space with velocity $v$ )

Einstein’s equation of photoemission:

h f = W_e + \frac{1}{2} m_e v_e^2

h f = W_e + e U

W_e = h f_0 \quad W_e = \text{work function} \quad f_0 = \text{threshold frequency}

e U = \text{voltage} \quad e = 1.602 \cdot 10^{-19} \, C

Franck–Hertz experiment proves that atoms absorb energy in quanta.

2. Calculate the energy of a photon corresponding to the extreme values of visible light. Violet has a wavelength λ_V=390 nm, red λ_R=790 nm. Express in joules and in eV. 1eV=1.602·10^-19J.

Solution:

Analysis:

A violet light photon has energy 3.18 eV, red 1.56 eV.

3. Compare the energy of a photon of yellow monochromatic light (λ = 500·10^-9m) with the average kinetic energy of a molecule of an ideal gas in its random motion at a temperature of 0⁰C.

Solution:

Analysis:

The energy of a yellow light photon is 70 times greater than the energy of a molecule of an ideal gas at 0⁰C.

4.The human eye can perceive light if the power of the light radiation incident on the eye is at least P=2·10^-17W. Determine how many photons with wavelength λ = 500·10^-9m hit the eye in 1 second.

Solution:

Analysis:

In one second, 50 photons hit the eye.

5.What is the threshold frequency of electromagnetic radiation needed to irradiate the surface of nickel so that the external photoelectric effect occurs? The work function of electrons from nickel is 5 eV.

Solution:

Analysis:

The threshold frequency of electromagnetic radiation for nickel is f₀ = 1.21·10¹⁵Hz. This is ultraviolet radiation.

6.Determine whether photoemission can occur when light with a wavelength λ = 390·10^-9m falls on zinc.

Work function for zinc is W_e = 4eV, λ = 390nm, W_e = 4eV = 4·1.602·10^-19J = 6.408·10^-19J, λ₀ = ?

Solution:

Analysis:

The photoelectric effect will not occur because λ₀ < λ.

7.At what speed do electrons leave a platinum plate (f₀ = 12.8·10¹⁴Hz), if ultraviolet light with wavelength λ = 150nm falls on it? (m_e = 9.1·10^-31kg)

Solution:

Analysis:

The electrons leave the platinum plate with a speed of about 10⁶m·s^-1.

8.The de Broglie wavelength of an accelerated electron is λ = 3.87·10^-11m. The electron was accelerated from rest in an electric field with voltage U. Calculate:

a) the speed of the electron
b) the accelerating voltage

Solution:

Analysis:

The electron moves at a speed of 1.9·10⁷m·s^-1
The accelerating voltage is 1kV

9.A photon of ultraviolet radiation has a wavelength λ = 100nm. Calculate

a) how many photons have an energy of 1J
b) how many photons have a mass of 1 microgram

Solution:

Analysis:

1J of energy corresponds to 5·10¹⁷ photons
A mass of 1μg corresponds to 4.5·10²⁵ photons

10.In the Compton experiment, the incident photon has frequency f₁=1.5·10²⁰Hz, the photon after collision has frequency f₂=1.1·10²⁰Hz.

a) Determine the energy gained by the electron that interacted with the photon. Express in eV
b) Determine the change in wavelength of the photon.

Solution:

Analysis:

The electron gains energy of 1.65 eV
The photon changes wavelength by 0.727·10^-12m

11.In the Franck–Hertz experiment, the authors found that the current drop occurs at a voltage in the electric field accelerating the electrons equal to U = 4.9V and that mercury vapors emit radiation with wavelength λ = 253.2nm. Calculate the value of Planck’s constant h.

Solution:

Analysis:

Planck’s constant is h = 6.625·10^–34J·s.

12. In the Compton effect of photon scattering on an electron, the photon wavelength changes by Δλ=4.85·10^–12m. By what angle υ does the photon deviate from the original direction?

Solution:

Analysis:

The scattering angle is υ = 180⁰. This is the so-called “backscattering”.

Physics

Quantum optics

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