Chapter 22 Quantum Physics
22.1 Energy and Momentum of a Photon
Wave and particulate nature of electromagnetic radiation
Electromagnetic radiation behaves as waves.
Evidence: EM radiation exhibits phenomenon like diffraction and interference.
Electromagnetic radiation behaves as particles.
Evidence: The photoelectric effect & the atomic line spectra.
Photons
The energy carried by electromagnetic radiation exists as discrete packets or quantities called quanta.
Photon
A “packet of energy” or a quantum of electromagnetic energy.
All electromagnetic radiation is consists of a stream of energy packets called photons.
Wave Theory of Light Concept of Quanta
▪ Suggests that the flow of electron in ▪ Suggests that the energy in light is
light waves is continuous. not transferred continuously, but as
discrete packets of energy.
▪ Each photon carries a specific
amount of energy and transfers this
energy all in one go, rather than
supplying a consistent amount of
energy.
▪ Gives rise to a particular nature of
electromagnetic waves.
Energy of a Photon
Einstein Relation
ℎ𝑐
𝐸 = ℎ𝑓 𝐸=
𝜆
E = energy of a photon of an EM radiation / J
h = Planck constant, 6.63 × 10-34 Js
f = frequency of the EM radiation / Hz
Energy of 1 photon = hf
c = speed of light, 3.00 × 108 ms-1
Energy of N photons = Nhf
λ = wavelength of the EM radiation / m
,E∝f
▪ The energy of a photon is directly proportional to the frequency of the electromagnetic waves.
▪ High frequency radiation means high-energy photons.
𝟏
E∝
𝝀
▪ The energy of a photon is inversely proportional to the wavelength.
▪ Short-wavelegth X-ray photon is far more energetic than long-wavelength photon of light.
The energy of a photon of visible light: 5.7 × 10-19 – 2.7 × 10-19 J
The energy of a photon is extremely small and far less than a joule.
The electronvolt (eV) is used as the energy unit.
Electronvolt (eV)
The energy gained by an electron when it is accelerated from rest in a vacuum through a potential
difference of one volt.
𝑄
𝑉=
𝑊
Energy change, 𝑊 = 𝑄𝑉
𝑊 = (1.60 × 10−19 )(1)
= 1.60 × 10−19 𝐽
𝟏𝒆𝑽 = 𝟏. 𝟔𝟎 × 𝟏𝟎−𝟏𝟗 𝑱
, Momentum of a Photon
The particle-like behaviour of photons.
Momentum, of a photon is related to its energy, E by the equation:
𝐸
𝜌=
𝑐
= momentum of a photon / Ns or kg ms-1
E = energy of photon / J
C: speed of light in a vacuum
Estimating the pressure exerted by photons hitting a metal plate
Eg: A 2.0 mW laser beam is incident normally on a fixed metal plate. The cross-sectional area of the beam is
4.0 × 10-6 m2. The light from the laser has frequency 4.7 × 1014 Hz. Calculate the momentum of the photon,
and the pressure exerted by the laser bream on the metal plate. Assume all photons are absorbed by the
plate.
1) Calculate the momentum of each photon.
𝐸 ℎ𝑓 6.63 × 10−34 × 4.7 × 1014
𝜌= = = = 1.04 × 10−27 𝑁𝑠
𝑐 𝑐 3.0 × 108
2) Calculate the no. of photons incident on the plate per second.
𝑃𝑜𝑤𝑒𝑟
𝑁𝑜. 𝑜𝑓 𝑝ℎ𝑜𝑡𝑜𝑛𝑠 𝑝𝑒𝑟 𝑠𝑒𝑐𝑜𝑛𝑑 =
𝐸𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑝ℎ𝑜𝑡𝑜𝑛
0.002 0.002
= = 6.63×10−34 ×4.7×1014 = 6.42 × 1015 𝑠 −1
ℎ𝑓
3) Calculate the force exerted on the plate by using Newton’s second law.
Consider a time interval of 1.0 s.
𝑓𝑜𝑟𝑐𝑒 = 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑜𝑓 𝑝ℎ𝑜𝑡𝑜𝑛
= 𝑛𝑜. 𝑜𝑓 𝑝ℎ𝑜𝑡𝑜𝑛𝑠 𝑝𝑒𝑟 𝑠𝑒𝑐𝑜𝑛𝑑 × 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑝ℎ𝑜𝑡𝑜𝑛
= 6.42 × 1015 × 1.04 × 10−27 = 6.68 × 10−12 𝑁
4) Calculate pressure.
𝐹 6.68 × 10−12
𝑃= = = 1.7 × 10−6 𝑃𝑎
𝐴 4.0 × 10−6
22.1 Energy and Momentum of a Photon
Wave and particulate nature of electromagnetic radiation
Electromagnetic radiation behaves as waves.
Evidence: EM radiation exhibits phenomenon like diffraction and interference.
Electromagnetic radiation behaves as particles.
Evidence: The photoelectric effect & the atomic line spectra.
Photons
The energy carried by electromagnetic radiation exists as discrete packets or quantities called quanta.
Photon
A “packet of energy” or a quantum of electromagnetic energy.
All electromagnetic radiation is consists of a stream of energy packets called photons.
Wave Theory of Light Concept of Quanta
▪ Suggests that the flow of electron in ▪ Suggests that the energy in light is
light waves is continuous. not transferred continuously, but as
discrete packets of energy.
▪ Each photon carries a specific
amount of energy and transfers this
energy all in one go, rather than
supplying a consistent amount of
energy.
▪ Gives rise to a particular nature of
electromagnetic waves.
Energy of a Photon
Einstein Relation
ℎ𝑐
𝐸 = ℎ𝑓 𝐸=
𝜆
E = energy of a photon of an EM radiation / J
h = Planck constant, 6.63 × 10-34 Js
f = frequency of the EM radiation / Hz
Energy of 1 photon = hf
c = speed of light, 3.00 × 108 ms-1
Energy of N photons = Nhf
λ = wavelength of the EM radiation / m
,E∝f
▪ The energy of a photon is directly proportional to the frequency of the electromagnetic waves.
▪ High frequency radiation means high-energy photons.
𝟏
E∝
𝝀
▪ The energy of a photon is inversely proportional to the wavelength.
▪ Short-wavelegth X-ray photon is far more energetic than long-wavelength photon of light.
The energy of a photon of visible light: 5.7 × 10-19 – 2.7 × 10-19 J
The energy of a photon is extremely small and far less than a joule.
The electronvolt (eV) is used as the energy unit.
Electronvolt (eV)
The energy gained by an electron when it is accelerated from rest in a vacuum through a potential
difference of one volt.
𝑄
𝑉=
𝑊
Energy change, 𝑊 = 𝑄𝑉
𝑊 = (1.60 × 10−19 )(1)
= 1.60 × 10−19 𝐽
𝟏𝒆𝑽 = 𝟏. 𝟔𝟎 × 𝟏𝟎−𝟏𝟗 𝑱
, Momentum of a Photon
The particle-like behaviour of photons.
Momentum, of a photon is related to its energy, E by the equation:
𝐸
𝜌=
𝑐
= momentum of a photon / Ns or kg ms-1
E = energy of photon / J
C: speed of light in a vacuum
Estimating the pressure exerted by photons hitting a metal plate
Eg: A 2.0 mW laser beam is incident normally on a fixed metal plate. The cross-sectional area of the beam is
4.0 × 10-6 m2. The light from the laser has frequency 4.7 × 1014 Hz. Calculate the momentum of the photon,
and the pressure exerted by the laser bream on the metal plate. Assume all photons are absorbed by the
plate.
1) Calculate the momentum of each photon.
𝐸 ℎ𝑓 6.63 × 10−34 × 4.7 × 1014
𝜌= = = = 1.04 × 10−27 𝑁𝑠
𝑐 𝑐 3.0 × 108
2) Calculate the no. of photons incident on the plate per second.
𝑃𝑜𝑤𝑒𝑟
𝑁𝑜. 𝑜𝑓 𝑝ℎ𝑜𝑡𝑜𝑛𝑠 𝑝𝑒𝑟 𝑠𝑒𝑐𝑜𝑛𝑑 =
𝐸𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑝ℎ𝑜𝑡𝑜𝑛
0.002 0.002
= = 6.63×10−34 ×4.7×1014 = 6.42 × 1015 𝑠 −1
ℎ𝑓
3) Calculate the force exerted on the plate by using Newton’s second law.
Consider a time interval of 1.0 s.
𝑓𝑜𝑟𝑐𝑒 = 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑜𝑓 𝑝ℎ𝑜𝑡𝑜𝑛
= 𝑛𝑜. 𝑜𝑓 𝑝ℎ𝑜𝑡𝑜𝑛𝑠 𝑝𝑒𝑟 𝑠𝑒𝑐𝑜𝑛𝑑 × 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑝ℎ𝑜𝑡𝑜𝑛
= 6.42 × 1015 × 1.04 × 10−27 = 6.68 × 10−12 𝑁
4) Calculate pressure.
𝐹 6.68 × 10−12
𝑃= = = 1.7 × 10−6 𝑃𝑎
𝐴 4.0 × 10−6
