Photons Class 12 Physics (NEB) Chapter 21 | Quantum Nature of Radiation, Photoelectric Equation, Stopping Potential, Planck’s Constant, Notes & PDF
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1. Syllabus

Unit: Modern Physics

Chapter 21: Photons (4 hours)

21.1 Quantum nature of radiation

21.2 Einstein’s photoelectric equation; stopping potential

21.3 Measurement of Planck’s constant

These notes follow the NEB Class 12 Physics syllabus (+2 Physics, Nepal). Read the quick overview first, practise the multiple choice questions, then work through the numericals, and finish with the PDF notes for quick revision.

2. Photons at a Glance

A photon is a quantum, or discrete packet, of electromagnetic radiation. According to the quantum nature of radiation (Planck 1900, Einstein 1905), radiation is emitted, travels and is absorbed in photons, not as a continuous wave of energy.

Properties and formulas of a photon

QuantityFormula or value
Energy of a photon\(E=hf=\dfrac{hc}{\lambda}\)
Momentum of a photon\(p=\dfrac h\lambda=\dfrac Ec\)
Planck’s constant\(h=6.63\times10^{-34}\ \mathrm{J\,s}\)
Useful product\(hc\approx1240\ \mathrm{eV\,nm}\), so \(E(\mathrm{eV})=\dfrac{1240}{\lambda(\mathrm{nm})}\)
Charge and rest massBoth zero; speed \(c\) in vacuum
Intensity of a beam\(I=\dfrac{nhf}{A}\), with \(n\) the number of photons per second

Photoelectric effect: laws, Einstein’s equation and stopping potential

The photoelectric effect is the emission of electrons from a metal surface when light of high enough frequency falls on it. Experiment shows that (1) there is a threshold frequency \(f_0\) below which no emission occurs, (2) the maximum kinetic energy of the electrons depends on the frequency, not on the intensity, (3) the photocurrent is proportional to the intensity, and (4) emission is instantaneous. The wave theory cannot explain these facts; Einstein’s photon theory does.

\[ hf=\phi+K_{\max},\qquad\phi=hf_0,\qquad eV_s=K_{\max}=hf-\phi \]
f Kmax f0 f0‘ −φ metal 1 metal 2 (smaller φ) slope = h (same for all metals)

Figure: Maximum kinetic energy against frequency. The slope is Planck’s constant \(h\); the intercept on the energy axis is \(-\phi\) and the intercept on the frequency axis is \(f_0\).

Measurement of Planck’s constant (Millikan)

Monochromatic light of known frequency falls on a metal cathode in a vacuum tube. The retarding voltage is increased until the photocurrent just becomes zero, which gives the stopping potential \(V_s\). From \(eV_s=hf-\phi\), a graph of \(V_s\) against \(f\) is a straight line of slope \(\dfrac he\). Planck’s constant is \(h=e\times\text{slope}\).

3. Multiple Choice Questions

Question 1. Einstein’s photoelectric equation states that \(E_k=E-W\). In this equation, \(E_k\) refers to:

  • a) Kinetic energy of all the emitted electrons
  • b) Mean kinetic energy of the emitted electrons
  • c) Maximum kinetic energy of the emitted electrons
  • d) Minimum kinetic energy of the emitted electrons

Correct Answer: c) Maximum kinetic energy of the emitted electrons

Reason: \(E\) is the photon energy and \(W\) is the work function, the minimum energy needed to eject an electron. An electron that loses no further energy on its way out leaves with the whole excess, so \(E_k\) is the maximum kinetic energy \(K_{\max}\).


Question 2. Emission of electrons from the surface of a metal by the action of light on it is:

  • a) Thermionic emission
  • b) Electronic emission
  • c) Photoelectric emission
  • d) Cold emission

Correct Answer: c) Photoelectric emission

Reason: Thermionic emission is caused by heat and cold (field) emission by a strong electric field. Emission caused by light falling on the metal is photoelectric emission.


Question 3. Which of the following radiations of the same intensity gives the maximum photocurrent?

  • a) Infra-red
  • b) Visible
  • c) Ultra-violet
  • d) All of the above

Correct Answer: a) Infra-red

Reason: Intensity is \(I=n\,hf\) per unit area, so for the same intensity the number of photons per second is \(n=\dfrac{I}{hf}\), which is largest for the lowest frequency. Infra-red has the lowest frequency, so it has the most photons. If every beam can eject electrons (frequency above the threshold) and each photon ejects one electron, infra-red gives the largest photocurrent.


Question 4. A photon is a:

  • a) Quantum of light energy
  • b) Quantum of matter
  • c) Positively charged particle
  • d) Instrument for measuring light intensity

Correct Answer: a) Quantum of light energy

Reason: By the Planck-Einstein relation \(E=hf\), electromagnetic radiation is quantised, and one packet of energy is called a photon.


Question 5. The photoelectric current can be increased by:

  • a) Increasing frequency
  • b) Increasing intensity
  • c) Decreasing intensity
  • d) Decreasing wavelength

Correct Answer: b) Increasing intensity

Reason: The photocurrent is proportional to the number of photoelectrons emitted per second. A greater intensity (with the frequency above the threshold) means more photons per second and so more emitted electrons.


Question 6. The study of the photoelectric effect is useful in understanding:

  • a) Quantisation of energy
  • b) Quantisation of charge
  • c) Conservation of charge
  • d) Conservation of kinetic energy

Correct Answer: a) Quantisation of energy

Reason: Classical wave theory could not explain the threshold frequency or the instantaneous emission. Einstein explained them by taking light energy to come in discrete packets, the photons, which establishes the quantisation of energy.

Question 7. A metal with a work function of 0.6 eV is illuminated with light of photon energy 2 eV. The stopping potential is:

  • a) 4.6 V
  • b) 0.8 V
  • c) 1.4 V
  • d) 3.6 V

Correct Answer: c) 1.4 V

\[ K_{\max}=E-\phi=2-0.6=1.4\ \mathrm{eV},\qquad eV_s=K_{\max}\;\Rightarrow\;V_s=1.4\ \mathrm{V} \]

Question 8. The work function of a metal is 2.51 eV. Its threshold frequency is:

  • a) \(7.5\times10^{13}\ \mathrm{Hz}\)
  • b) \(6.08\times10^{14}\ \mathrm{Hz}\)
  • c) \(10.4\times10^{12}\ \mathrm{Hz}\)
  • d) \(6.8\times10^{11}\ \mathrm{Hz}\)

Correct Answer: b) \(6.08\times10^{14}\ \mathrm{Hz}\)

\[ \phi=2.51\times1.6\times10^{-19}=4.016\times10^{-19}\ \mathrm{J},\qquad f_0=\frac\phi h=\frac{4.016\times10^{-19}}{6.6\times10^{-34}}\approx6.08\times10^{14}\ \mathrm{Hz} \]

Question 9. The threshold frequency of a metal is \(f_0\). When light of frequency \(4f_0\) is incident on the metal, the maximum kinetic energy of the emitted photoelectrons is:

  • a) \(5hf_0\)
  • b) \(hf_0/2\)
  • c) \(hf_0\)
  • d) \(3hf_0\)

Correct Answer: d) \(3hf_0\)

\[ K_{\max}=h(4f_0)-\phi=4hf_0-hf_0=3hf_0 \]

Question 10. Ultraviolet radiation of 6.2 eV falls on an aluminium surface (work function 4.2 eV). The kinetic energy in joules of the fastest electron is approximately:

  • a) \(5.2\times10^{-17}\ \mathrm{J}\)
  • b) \(3\times10^{-19}\ \mathrm{J}\)
  • c) \(2.4\times10^{-18}\ \mathrm{J}\)
  • d) \(2.65\times10^{-16}\ \mathrm{J}\)

Correct Answer: b) \(3\times10^{-19}\ \mathrm{J}\)

\[ K_{\max}=6.2-4.2=2.0\ \mathrm{eV}=2.0\times1.6\times10^{-19}=3.2\times10^{-19}\ \mathrm{J} \]

The nearest option is \(3\times10^{-19}\ \mathrm{J}\).

4. Numerical and Subjective Questions

Question 1. Threshold frequency, stopping potential and work function.

a) Define threshold frequency, stopping potential and work function in photoelectric emission.

Threshold frequency \(f_0\): the minimum frequency of the incident radiation that can just eject photoelectrons from a given metal surface. Below it no emission occurs, however intense the light.

Stopping potential \(V_s\): the minimum retarding (negative) potential of the anode with respect to the cathode that stops even the most energetic photoelectrons, reducing the photocurrent to zero.

Work function \(\phi\): the minimum energy needed by an electron to just escape from the metal surface; \(\phi=hf_0\).

b) Show that the existence of a threshold frequency is evidence in favour of the quantum theory.

In the classical wave theory the energy is spread continuously over the wavefront, so a low-frequency but intense (or long-lasting) wave should eventually give an electron enough accumulated energy to escape. The wave theory therefore predicts no threshold frequency.

In the quantum theory light consists of photons of energy \(E=hf\), and one electron absorbs one photon. If \(hf<\phi\) the electron cannot escape, no matter how many photons arrive per second or for how long. The observed sharp threshold frequency fits this one-photon, one-electron interaction.


Question 2. In the photoelectric effect, electrons leave a metal surface when electromagnetic radiation strikes it. When radiation of less than a certain frequency strikes the surface, there is no emission of electrons.

a) (i) Explain why there is no emission if the frequency is lower than a certain value.

The radiation arrives as photons of energy \(E=hf\). An electron needs at least the work function \(\phi\) to leave the metal. If \(f\) is below the threshold, \(hf<\phi\), and since an electron absorbs only one photon at a time it never gets enough energy to escape.

(ii) State two other pieces of evidence from the photoelectric effect that suggest electromagnetic radiation has particle properties.

Instantaneous emission: electrons are emitted almost at once (in less than about \(10^{-9}\ \mathrm{s}\)) when light strikes the surface, with none of the time delay for energy to build up that wave theory predicts.

Kinetic energy independent of intensity: the maximum kinetic energy of the electrons depends only on the frequency of the light, not on its intensity.

b) (i) The work function of the metal is 3.8 eV. Calculate the minimum frequency of radiation that causes photoelectric emission.

\[ \phi=3.8\times1.6\times10^{-19}=6.08\times10^{-19}\ \mathrm{J},\qquad f_0=\frac\phi h=\frac{6.08\times10^{-19}}{6.63\times10^{-34}}\approx9.17\times10^{14}\ \mathrm{Hz} \]

(ii) An electron at the surface of the metal is emitted with a kinetic energy of \(4.5\times10^{-19}\ \mathrm{J}\). Calculate the energy of the incident photon in eV.

\[ K=\frac{4.5\times10^{-19}}{1.6\times10^{-19}}=2.81\ \mathrm{eV},\qquad E=K+\phi=2.81+3.8\approx6.61\ \mathrm{eV} \]

c) Mention any two applications of the photoelectric effect in daily life.

Solar (photovoltaic) cells, which convert sunlight directly into electrical energy; and light meters in photography, which measure light intensity. (Automatic doors and burglar alarms also use photocells.)


Question 3. Millikan’s experiment and a potassium surface.

a) Describe Millikan’s experiment to calculate the value of Planck’s constant.

Millikan used a vacuum tube with an alkali-metal cathode and an anode. Monochromatic light of several known frequencies \(f\) was shone on the cathode. For each frequency a variable retarding voltage was applied to the anode until the photocurrent just became zero, giving the stopping potential \(V_s\).

From Einstein’s equation, \(eV_s=hf-\phi\), so

\[ V_s=\frac he\,f-\frac\phi e \]

A graph of \(V_s\) against \(f\) is a straight line of slope \(\dfrac he\). Multiplying the measured slope by the electron charge \(e\) gives Planck’s constant \(h\).

b) The photoelectric work function of potassium is \(3.8\times10^{-19}\ \mathrm{J}\). If light of wavelength 5000 \(\mathrm{\AA}\) is incident on it, find (i) the stopping potential, (ii) the kinetic energy of the most energetic electrons and (iii) the velocity of these electrons.

Given: \(\phi=3.8\times10^{-19}\ \mathrm{J},\ \lambda=5\times10^{-7}\ \mathrm{m},\ h=6.62\times10^{-34}\ \mathrm{J\,s},\ c=3\times10^8\ \mathrm{m\,s^{-1}},\ m=9.1\times10^{-31}\ \mathrm{kg}\)
\[ E=\frac{hc}{\lambda}=\frac{6.62\times10^{-34}\times3\times10^8}{5\times10^{-7}}=3.972\times10^{-19}\ \mathrm{J} \]

(ii) Kinetic energy:

\[ K_{\max}=E-\phi=3.972\times10^{-19}-3.8\times10^{-19}=1.72\times10^{-20}\ \mathrm{J} \]

(i) Stopping potential:

\[ V_s=\frac{K_{\max}}{e}=\frac{1.72\times10^{-20}}{1.6\times10^{-19}}=0.1075\ \mathrm{V} \]

(iii) Velocity:

\[ v=\sqrt{\frac{2K_{\max}}{m}}=\sqrt{\frac{2\times1.72\times10^{-20}}{9.1\times10^{-31}}}\approx1.94\times10^{5}\ \mathrm{m\,s^{-1}} \]
Question 4. Einstein’s photoelectric equation and its laws.

a) What is the photoelectric effect? Obtain Einstein’s photoelectric equation. State and explain the laws of photoelectric emission from this equation.

Photoelectric effect: the emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequency falls on it.

Einstein’s equation: light consists of photons of energy \(E=hf\). A photon gives its whole energy to a single electron. Part of it, the work function \(\phi\), is used to overcome the surface barrier, and the rest appears as the maximum kinetic energy of the emitted electron. By conservation of energy

\[ hf=\phi+K_{\max}\;\Rightarrow\;K_{\max}=hf-\phi=\tfrac12mv_{\max}^2 \]

Laws from the equation:

Threshold frequency: emission needs \(K_{\max}\ge0\), that is \(hf\ge\phi\), so there is a minimum frequency \(f_0=\phi/h\) below which no electrons are emitted.

Independence of intensity: \(K_{\max}\) depends only on \(f\) and \(\phi\). A greater intensity means more photons per second, hence a larger photocurrent, but not a larger \(K_{\max}\).

Dependence on frequency: \(K_{\max}\) increases linearly with the frequency of the light.

b) A sheet of silver is illuminated by monochromatic ultraviolet radiation of wavelength 1810 \(\mathrm{\AA}\). What is the maximum energy of the emitted electron? The threshold wavelength of silver is 2640 \(\mathrm{\AA}\).

Using \(hc\approx1240\ \mathrm{eV\,nm}\), with \(\lambda=181\ \mathrm{nm}\) and \(\lambda_0=264\ \mathrm{nm}\):

\[ K_{\max}=hc\left(\frac1\lambda-\frac1{\lambda_0}\right)=1240\left(\frac1{181}-\frac1{264}\right)\approx2.15\ \mathrm{eV} \]

Question 5. Einstein’s theory and photosynthesis.

a) Explain Einstein’s theory of photoelectricity. Hence arrive at Einstein’s photoelectric equation.

Einstein postulated that light interacts with matter in quanta. An electron near the surface absorbs one whole photon of energy \(hf\). If \(hf>\phi\), the electron uses \(\phi\) to break free of the binding forces and keeps the excess as kinetic energy:

\[ hf=\phi+\tfrac12mv_{\max}^2 \]

This is Einstein’s photoelectric equation (see Question 4a).

b) Photosynthesis starts in certain plants when they are exposed to sunlight but not if the plant is exposed only to infrared light. Explain.

Photosynthesis is a photochemical reaction, and each molecule that starts the reaction must absorb a single photon with at least a certain minimum energy. Sunlight contains visible and ultraviolet light, whose photons have a high frequency and energy \(hf\) large enough to trigger the reaction. Infrared photons have a lower frequency and so too little energy per photon. Since a molecule absorbs one photon at a time, the infrared photons cannot start the process, however intense the infrared light is.


Question 6. A parallel beam of monochromatic light of wavelength 660 nm has an intensity of 0.25 mW m\(^{-2}\). The beam is incident normally on a surface where the light is totally absorbed. For a cross-sectional area of 1.2 cm\(^2\) of the beam, calculate (a) the number of photons passing per second, (b) the momentum of a photon and (c) the force exerted on the surface.
Given: \(\lambda=660\ \mathrm{nm},\ I=0.25\times10^{-3}\ \mathrm{W\,m^{-2}},\ A=1.2\times10^{-4}\ \mathrm{m^2},\ h=6.63\times10^{-34}\ \mathrm{J\,s},\ c=3.00\times10^8\ \mathrm{m\,s^{-1}}\)

(a) Photons per second: the power through the area is

\[ P=IA=0.25\times10^{-3}\times1.2\times10^{-4}=3.0\times10^{-8}\ \mathrm{W} \]

The energy of one photon is

\[ E=\frac{hc}{\lambda}=\frac{6.63\times10^{-34}\times3.00\times10^{8}}{660\times10^{-9}}=3.01\times10^{-19}\ \mathrm{J} \]
\[ n=\frac PE=\frac{3.0\times10^{-8}}{3.01\times10^{-19}}\approx1.0\times10^{11}\ \mathrm{s^{-1}} \]

(b) Momentum of a photon:

\[ p=\frac h\lambda=\frac{6.63\times10^{-34}}{660\times10^{-9}}\approx1.0\times10^{-27}\ \mathrm{N\,s} \]

(c) Force: the light is totally absorbed, so each photon loses its whole momentum \(p\) and the force is the rate of change of momentum:

\[ F=np=1.0\times10^{11}\times1.0\times10^{-27}=1.0\times10^{-16}\ \mathrm{N} \]

Question 7. Electrons are emitted from a metal surface when it is illuminated with suitable electromagnetic radiation. The variation with frequency \(f\) of the maximum kinetic energy \(E_k\) of the emitted electrons is shown in the figure.
Class 12 Physics Photons graph of maximum kinetic energy of photoelectrons against frequency

Figure: Maximum kinetic energy of the photoelectrons against frequency

a) Name the effect described above.

The photoelectric effect.

b) Use the figure to determine (i) the threshold frequency of the radiation and (ii) a value for Planck’s constant.

(i) Threshold frequency: from \(E_k=hf-\phi\), \(E_k=0\) at \(f=f_0\), so \(f_0\) is the intercept of the graph on the frequency axis. From the figure, \(f_0=6.8\times10^{14}\ \mathrm{Hz}\).

(ii) Planck’s constant: the gradient of the straight line is \(h\). Taking the gradient of the line in the figure,

\[ h\approx6.62\times10^{-34}\ \mathrm{J\,s} \]

c) On the figure draw a line to show the variation of \(E_k\) with \(f\) for a second metal which has a lower work function than the metal in (b).

A lower work function means a lower threshold frequency, \(f_0=\phi/h\). The gradient is still \(h\), so the new line must be a straight line parallel to the original, shifted to the left, so that it meets the frequency axis at a smaller value of \(f\).

d) The kinetic energy of the electrons is described as the maximum. Suggest why the emitted electrons are likely to have a range of kinetic energies for any one frequency.

Einstein’s equation gives the energy of electrons that come from the very surface of the metal and lose nothing on the way out. Electrons from deeper in the metal need more than \(\phi\) to reach the surface, because they lose varying amounts of energy in collisions with atoms. They therefore leave with a range of kinetic energies from zero up to the maximum value.


Question 8. A stationary isolated nucleus emits a \(\gamma\)-ray photon of energy 0.51 MeV.

a) State what is meant by a photon.

A photon is a discrete packet, or quantum, of electromagnetic radiation energy.

b) For the \(\gamma\)-ray photon, calculate (i) its wavelength and (ii) its momentum.

\(E=0.51\ \mathrm{MeV}=0.51\times10^6\times1.6\times10^{-19}=8.16\times10^{-14}\ \mathrm{J}\)

(i) Wavelength:

\[ \lambda=\frac{hc}{E}=\frac{6.63\times10^{-34}\times3\times10^8}{8.16\times10^{-14}}\approx2.4\times10^{-12}\ \mathrm{m} \]

(ii) Momentum:

\[ p=\frac Ec=\frac{8.16\times10^{-14}}{3\times10^8}\approx2.7\times10^{-22}\ \mathrm{N\,s} \]

c) For this nucleus, determine the change in mass \(\Delta m\) during the decay that gives rise to the energy of the \(\gamma\)-ray photon.

\[ \Delta m=\frac{E}{c^2}=\frac{8.16\times10^{-14}}{(3\times10^8)^2}\approx9.1\times10^{-31}\ \mathrm{kg} \]

Question 9. A mercury vapour lamp, caesium and tungsten.

a) Explain what is meant by the photoelectric effect.

The photoelectric effect is the ejection of electrons from the surface of a material, usually a metal, when electromagnetic radiation of sufficiently high frequency falls on it.

b) One wavelength of electromagnetic radiation emitted from a mercury vapour lamp is 436 nm. Calculate the photon energy corresponding to this wavelength.

\[ E=\frac{hc}{\lambda}=\frac{6.63\times10^{-34}\times3.0\times10^{8}}{436\times10^{-9}}\approx4.56\times10^{-19}\ \mathrm{J} \]

c) The light from the lamp falls, separately, on caesium (work function 1.4 eV) and tungsten (work function 4.5 eV). Calculate the threshold wavelength for photoelectric emission from (i) caesium and (ii) tungsten.

Use \(\phi=\dfrac{hc}{\lambda_0}\), so \(\lambda_0=\dfrac{hc}{\phi}\) with \(hc\approx1240\ \mathrm{eV\,nm}\):

\[ \lambda_{0,\mathrm{Cs}}=\frac{1240}{1.4}\approx886\ \mathrm{nm},\qquad\lambda_{0,\mathrm{W}}=\frac{1240}{4.5}\approx276\ \mathrm{nm} \]

d) Use your answers in (c) to state and explain whether the radiation of wavelength 436 nm will cause photoelectric emission from each of the metals.

Emission occurs only if the incident wavelength is less than or equal to the threshold wavelength, that is, if the photon energy is at least the work function.

Caesium: \(436\ \mathrm{nm}<886\ \mathrm{nm}\), so the photon energy (about 2.8 eV) exceeds 1.4 eV and photoelectric emission occurs.

Tungsten: \(436\ \mathrm{nm}>276\ \mathrm{nm}\), so the photon energy (about 2.8 eV) is less than 4.5 eV and there is no photoelectric emission.

5. View PDF Notes

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6. Frequently Asked Questions

What is a photon in physics?
A photon is a quantum, or discrete packet, of electromagnetic radiation. It travels at the speed of light c in vacuum, has no charge and zero rest mass, and carries energy E = hf = hc/λ and momentum p = h/λ = E/c, where h = 6.63 × 10⁻³⁴ J s is Planck’s constant.
What are photons made of?
Photons are elementary particles, so they are not made of anything smaller. A photon is the quantum of the electromagnetic field: it has no electric charge, zero rest mass and spin 1, and it is the particle that carries the electromagnetic force.
Are photons energy?
A photon is a particle of light that carries energy, E = hf, together with momentum. So energy is a property of the photon rather than the photon being energy by itself. The more energetic the radiation (higher frequency, shorter wavelength), the more energy each photon carries.
What do photons do?
Photons carry electromagnetic energy and momentum from one place to another and mediate the electromagnetic force between charges. When absorbed, a photon can eject an electron from a metal (the photoelectric effect), excite chlorophyll in photosynthesis, produce a current in a solar cell or trigger vision in the eye.
What are photons of light?
Light is made of photons. A photon of visible light has an energy of roughly 1.7 eV (red, about 700 nm) to 3.1 eV (violet, about 400 nm). Photon energy in electron-volts can be found from E = 1240/λ with λ in nanometres. The intensity of a light beam is the number of photons crossing unit area per second multiplied by the energy of one photon.
What is the quantum nature of radiation?
The quantum nature of radiation means that electromagnetic radiation is emitted, travels and is absorbed in discrete packets of energy called quanta or photons, not as a continuous wave of energy. Planck introduced the idea in 1900 and Einstein used it in 1905 to explain the photoelectric effect. Radiation shows both wave nature (interference, diffraction) and particle nature (photoelectric effect).
What are the formulas for the quantum nature of radiation?
Energy of a photon: E = hf = hc/λ. Momentum of a photon: p = h/λ = E/c. Einstein’s photoelectric equation: hf = φ + Kmax, with φ = hf₀ the work function. Stopping potential: eVs = Kmax = hf – φ. Planck’s constant is h = 6.63 × 10⁻³⁴ J s and the useful product hc is about 1240 eV nm.
What is Planck’s quantum theory?
Planck’s quantum theory (1900) states that energy is emitted or absorbed by a body in discrete amounts called quanta, and the energy of one quantum of radiation of frequency f is E = hf, so the total energy exchanged is an integer multiple nhf. It explained the spectrum of black-body radiation, which classical physics could not.
What is the quantum theory of electromagnetic radiation?
The quantum theory of electromagnetic radiation describes the electromagnetic field as made of photons. Each photon has energy hf and momentum h/λ, and radiation is emitted and absorbed one photon at a time. At the advanced level it is developed as quantum electrodynamics, treated in textbooks such as Walter Heitler’s The Quantum Theory of Radiation.
What is the photoelectric effect and what are its laws?
The photoelectric effect is the emission of electrons from a metal surface when light of sufficiently high frequency falls on it. Its laws are: emission occurs only if the frequency is above a threshold frequency f₀ for the metal; the maximum kinetic energy of the electrons depends on the frequency and not on the intensity; the photocurrent is proportional to the intensity; and the emission is instantaneous.
What is Einstein’s photoelectric equation?
Einstein’s photoelectric equation is hf = φ + Kmax, or Kmax = hf – φ. A photon of energy hf is absorbed by one electron; the work function φ is the energy needed to leave the metal and the remaining energy is the maximum kinetic energy. A metal of work function 0.6 eV illuminated by 2 eV photons gives Kmax = 1.4 eV and a stopping potential of 1.4 V.
What are stopping potential and threshold frequency?
The stopping potential Vs is the minimum retarding potential that stops even the most energetic photoelectrons, so that the photocurrent becomes zero; eVs = Kmax. The threshold frequency f₀ is the minimum frequency of light that can cause photoelectric emission from a metal, f₀ = φ/h. The work function is the minimum energy needed to remove an electron from the metal.
How is Planck’s constant measured?
In Millikan’s experiment, light of several known frequencies falls on a metal cathode in a vacuum tube and the stopping potential Vs is found for each. From eVs = hf – φ, a graph of Vs against f is a straight line with slope h/e. Multiplying the slope by the electron charge e gives Planck’s constant h.
What are photons in photosynthesis?
In photosynthesis, photons of visible light, mainly red and blue, are absorbed by chlorophyll, and each absorbed photon gives an electron the energy needed to start the chemical reactions. Infrared photons have too little energy per photon to start the process, so a plant exposed only to infrared light does not photosynthesise, however intense the light is.
How is the word photon pronounced?
Photon is pronounced FOH-ton, with the stress on the first syllable (British /ˈfəʊtɒn/, American /ˈfoʊtɑːn/). It comes from the Greek word phos, meaning light.
What is a quantum of radiation? (crossword clue)
A quantum of radiation, the usual crossword answer, is a PHOTON (6 letters), the packet of electromagnetic energy E = hf.
What topics are included in Chapter 21 Photons of NEB Class 12 Physics?
The chapter (4 hours, part of the Modern Physics unit) covers: 21.1 quantum nature of radiation; 21.2 Einstein’s photoelectric equation and stopping potential; 21.3 measurement of Planck’s constant.
Where can I get Class 12 Physics Photons notes in PDF?
This page includes handwritten notes and typed PDF notes for Photons that you can view online, along with a graph of the photoelectric equation, solved multiple choice questions and numericals.

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