CBSE Class 12 Physics (2026–27)
Chapter 11: Dual Nature of Radiation and Matter
20 Important Questions and Answers
1. What is the dual nature of radiation?
Answer:
The dual nature of radiation means that light exhibits both wave and particle characteristics. Wave nature is evident from phenomena such as interference, diffraction, and polarization. However, experiments like the photoelectric effect show that light also behaves as a stream of particles called photons. Each photon carries a definite amount of energy given by (E = h\nu), where (h) is Planck’s constant and (\nu) is the frequency of light. Depending on the experiment being performed, light may display wave-like or particle-like properties. This concept was a major breakthrough in modern physics and formed the basis of quantum theory.
2. State Einstein’s photoelectric equation and explain its significance.
Answer:
Einstein’s photoelectric equation is:
[h\nu = \phi + K_{\max}]
where (h\nu) is the energy of the incident photon, (\phi) is the work function of the metal, and (K_{\max}) is the maximum kinetic energy of the emitted electron. The equation explains that a part of the photon’s energy is used to remove the electron from the metal surface, while the remaining energy appears as kinetic energy. This equation successfully explained experimental observations of the photoelectric effect that classical wave theory could not. It provided strong evidence for the particle nature of light and earned Einstein the Nobel Prize in Physics.
3. What is the photoelectric effect?
Answer:
The photoelectric effect is the phenomenon in which electrons are emitted from a metal surface when electromagnetic radiation of sufficiently high frequency falls on it. The emitted electrons are called photoelectrons. The effect occurs only when the frequency of incident light is greater than a certain minimum value known as the threshold frequency. The number of emitted electrons depends on the intensity of light, while their maximum kinetic energy depends on the frequency of light. The photoelectric effect could not be explained by classical wave theory but was successfully explained by Einstein using the photon concept.
4. What is threshold frequency?
Answer:
Threshold frequency is the minimum frequency of incident radiation required to eject electrons from a metal surface. It is represented by (\nu_0). If the frequency of the incident light is less than the threshold frequency, no photoelectric emission takes place regardless of the intensity of light. The threshold frequency depends on the nature of the metal and its work function. The relation between threshold frequency and work function is:
[\phi = h\nu_0]
where (\phi) is the work function and (h) is Planck’s constant. Metals with lower work functions have lower threshold frequencies and emit photoelectrons more easily.
5. Define work function of a metal.
Answer:
The work function of a metal is the minimum energy required to remove an electron from the surface of the metal. It is denoted by (\phi) and is usually measured in electron volts (eV). Different metals have different work functions depending on how strongly electrons are bound within them. According to Einstein’s photoelectric equation, the work function is related to the threshold frequency by:
[\phi = h\nu_0]
If the energy of the incident photon is less than the work function, photoelectric emission cannot occur. The concept of work function is important in understanding photoelectric devices and electron emission processes.
6. What are photons?
Answer:
Photons are the fundamental particles or quanta of electromagnetic radiation. According to Planck’s quantum theory and Einstein’s explanation of the photoelectric effect, light consists of discrete packets of energy called photons. The energy of a photon is given by:
[E = h\nu]
where (h) is Planck’s constant and (\nu) is the frequency of radiation. Photons have zero rest mass and always travel at the speed of light in vacuum. They possess momentum given by (p = h/\lambda). The photon concept successfully explains various phenomena such as the photoelectric effect, Compton effect, and blackbody radiation.
7. Why does photoelectric emission occur instantaneously?
Answer:
Photoelectric emission occurs instantaneously because a single photon transfers its entire energy to a single electron in one interaction. If the photon energy is greater than the work function of the metal, the electron is emitted immediately without any time delay. Classical wave theory predicted that electrons would gradually absorb energy and take time before emission. However, experiments showed that photoelectrons are emitted almost instantly, even at very low light intensities. Einstein’s photon theory explained this observation successfully and provided strong evidence that light energy is transferred in discrete packets rather than continuously.
8. Explain the effect of intensity of light on photoelectric current.
Answer:
The photoelectric current depends directly on the intensity of incident light, provided the frequency of light is greater than the threshold frequency. Higher intensity means more photons strike the metal surface per second. As a result, more electrons are emitted, producing a larger photoelectric current. However, the maximum kinetic energy of the emitted electrons remains unchanged because it depends only on the frequency of light and not on its intensity. Thus, increasing intensity increases the number of photoelectrons but does not affect their energy. This observation supports Einstein’s photon theory of light.
9. Explain the effect of frequency of light on photoelectric emission.
Answer:
The frequency of incident light plays a crucial role in photoelectric emission. If the frequency is below the threshold frequency, no photoelectrons are emitted regardless of the intensity. When the frequency exceeds the threshold value, photoelectric emission begins. As the frequency increases further, the maximum kinetic energy of emitted electrons also increases according to Einstein’s equation:
[K_{\max}=h\nu-\phi]
Thus, frequency determines the energy of photoelectrons, whereas intensity determines their number. This experimental observation contradicted classical wave theory and strongly supported the quantum theory of radiation proposed by Einstein.
10. What is stopping potential?
Answer:
Stopping potential is the minimum negative potential applied to the collector plate that just prevents the most energetic photoelectrons from reaching it. It is denoted by (V_0). At this potential, the photoelectric current becomes zero. The maximum kinetic energy of photoelectrons is related to stopping potential by:
[K_{\max}=eV_0]
where (e) is the electronic charge. Stopping potential depends on the frequency of incident light but is independent of its intensity. Measurement of stopping potential provides an experimental method for determining the kinetic energy of emitted photoelectrons.
11. State de Broglie hypothesis.
Answer:
According to de Broglie’s hypothesis, every moving particle possesses wave-like properties. Just as light exhibits both wave and particle nature, material particles such as electrons, protons, and atoms also behave like waves under suitable conditions. The wavelength associated with a moving particle is called the de Broglie wavelength and is given by:
[\lambda=\frac{h}{p}]
where (h) is Planck’s constant and (p) is the momentum of the particle. This revolutionary idea extended the concept of wave-particle duality to matter and laid the foundation for wave mechanics and quantum physics.
12. What is de Broglie wavelength?
Answer:
The de Broglie wavelength is the wavelength associated with a moving material particle. It is given by:
[\lambda=\frac{h}{mv}]
where (h) is Planck’s constant, (m) is the mass of the particle, and (v) is its velocity. The wavelength is inversely proportional to the momentum of the particle. For microscopic particles such as electrons, the wavelength is significant and can be observed experimentally. For macroscopic objects, the wavelength is extremely small and cannot be detected. The concept explains the wave nature of matter and has been verified through electron diffraction experiments.
13. Why is the de Broglie wavelength of macroscopic objects negligible?
Answer:
The de Broglie wavelength of a particle is given by:
[\lambda=\frac{h}{mv}]
For macroscopic objects, the mass is very large. Since wavelength is inversely proportional to mass, the resulting de Broglie wavelength becomes extremely small, often many orders of magnitude smaller than atomic dimensions. Such tiny wavelengths cannot be detected using available experimental techniques. Therefore, wave properties are not observable for everyday objects like balls, cars, or humans. However, for microscopic particles such as electrons and neutrons, the wavelength is appreciable and their wave behavior can be observed experimentally.
14. Write the expression for momentum of a photon.
Answer:
Although photons have zero rest mass, they possess momentum because they carry energy. The momentum of a photon is given by:
[p=\frac{h}{\lambda}]
where (h) is Planck’s constant and (\lambda) is the wavelength of light. Using the relation (E=h\nu), photon momentum can also be written as:
[p=\frac{E}{c}]
where (E) is photon energy and (c) is the speed of light. The existence of photon momentum explains phenomena such as radiation pressure and the Compton effect. It also supports the particle nature of electromagnetic radiation.
15. What experimental evidence supports de Broglie’s hypothesis?
Answer:
The strongest experimental evidence supporting de Broglie’s hypothesis came from the Davisson-Germer experiment. In this experiment, electrons were scattered by a nickel crystal and produced diffraction patterns similar to those produced by X-rays. Diffraction is a wave phenomenon, indicating that electrons possess wave properties. The experimentally measured wavelength matched the de Broglie wavelength predicted theoretically. Similar evidence was obtained through electron diffraction experiments by G. P. Thomson. These experiments confirmed that matter particles exhibit wave behavior and validated de Broglie’s theory.
16. Explain the Davisson-Germer experiment.
Answer:
The Davisson-Germer experiment demonstrated the wave nature of electrons. A beam of electrons accelerated through a known potential difference was directed onto a nickel crystal. The scattered electrons were detected at different angles. The intensity distribution showed diffraction maxima similar to X-ray diffraction patterns. Using Bragg’s law, the wavelength of electrons was calculated and found to agree with the de Broglie wavelength. This experiment provided direct evidence that electrons behave as waves. It confirmed de Broglie’s hypothesis and played an important role in the development of quantum mechanics.
17. Differentiate between photons and electrons.
Answer:
Photons are particles of electromagnetic radiation, while electrons are material particles. Photons have zero rest mass and always travel at the speed of light. Electrons possess finite rest mass and move with speeds less than that of light. The energy of a photon depends on its frequency, whereas an electron’s energy depends on its motion. Both photons and electrons exhibit wave-particle duality. Photons show wave behavior through interference and diffraction, while electrons display wave properties through diffraction experiments. Thus, both are quantum particles but differ in mass, charge, and origin.
18. What are the limitations of classical wave theory in explaining the photoelectric effect?
Answer:
Classical wave theory failed to explain several observations of the photoelectric effect. It could not explain the existence of a threshold frequency below which no emission occurs. It predicted that electron energy should increase with light intensity, whereas experiments showed dependence on frequency. The theory also suggested a time delay in electron emission, but emission occurs instantaneously. Furthermore, it could not explain why stopping potential depends on frequency rather than intensity. Einstein’s photon theory successfully explained all these observations by assuming that light energy is transferred in discrete packets called photons.
19. How does increasing accelerating potential affect de Broglie wavelength?
Answer:
When a charged particle such as an electron is accelerated through a potential difference (V), it gains kinetic energy. Its de Broglie wavelength is given by:
[\lambda=\frac{h}{\sqrt{2meV}}]
where (m) is the mass of the electron and (e) is its charge. Since wavelength is inversely proportional to the square root of accelerating potential, increasing the potential difference increases the electron’s momentum and decreases its wavelength. Thus, faster electrons have shorter de Broglie wavelengths. This principle is used in electron microscopes, where high-speed electrons provide very high resolution.
20. State the main conclusions of the dual nature of matter and radiation.
Answer:
The dual nature concept states that both radiation and matter exhibit wave as well as particle characteristics. Light behaves as a wave in interference and diffraction experiments but acts as photons in the photoelectric effect. Similarly, material particles such as electrons possess wave properties described by de Broglie wavelength. The observed nature depends on the experimental conditions. This principle led to the development of quantum mechanics and revolutionized modern physics. It explains the behavior of microscopic particles and forms the basis of technologies such as electron microscopes, semiconductor devices, lasers, and quantum electronics.