Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

11. 1 In Class 12 Physics, the chapter on "Dual Nature of Radiation and Matter" explores a fascinating concept: that light and matter can behave both as waves and as particles. This chapter introduces key ideas like the photoelectric effect, where light hitting a metal surface can release electrons, and de Broglie’s hypothesis, which suggests that particles like electrons also have wave-like properties. Understanding these dual characteristics helps explain various physical phenomena and bridges classical and quantum physics. By studying this chapter, you'll gain insight into how the behaviour of light and matter can be described in different ways, leading to a deeper comprehension of the natural world.

11.2 Electron Emission: Electron emission refers to the process where electrons are released from a material, often a metal, when it absorbs energy. This can happen in several ways, depending on the type of energy source:

1. Photoelectric Emission: When light (or photons) strikes a metal surface, it can cause the ejection of electrons. The energy of the photons must be greater than the work function (the minimum energy required to remove an electron from the metal). The formula for this is:

$E_k = h\nu - \phi $

where $ E_k $ is the kinetic energy of the emitted electron, h is Planck's constant, $\nu$ is the frequency of the incident light, and $ \phi $ is the work function of the metal.

2. Thermionic Emission: This occurs when a metal is heated to a high temperature, causing electrons to gain enough energy to overcome the work function and escape from the metal surface.

3. Field Emission: This happens when a strong electric field is applied to a material, which can pull electrons out of the surface.

These processes illustrate how energy input can lead to the emission of electrons, demonstrating the interaction between matter and energy in different ways.

11.3 Photoelectric Effect

Hertz’s observations: Hertz’s observations are key to understanding the photoelectric effect. Heinrich Hertz, a physicist, conducted experiments in the late 19th century that revealed how ultraviolet light could cause electrons to be emitted from a metal surface.

Hertz observed that when ultraviolet light struck a metal electrode, it emitted electrons, but this only happened if the light’s frequency was above a certain threshold. He found that:

1. The number of emitted electrons increased with the intensity of the light.

2. The energy of the emitted electrons depended on the frequency of the light, not its intensity.

These observations were crucial in showing that light must have enough energy to free electrons from the metal surface, leading to the development of the concept of photons.

Formula:

The energy of the emitted electrons is given by:$ E_k = h\nu - \phi$

Here:

$ E_k$ is the kinetic energy of the emitted electrons.

h is Planck’s constant.

$\nu $ is the frequency of the incident light.

$ \phi$ is the work function of the metal.

Hertz’s work laid the foundation for the quantum theory of light and helped in understanding the dual nature of radiation.

Hallwachs’ and Lenard’s observations:

Hallwachs’ Observation: In the late 19th century, Wilhelm Hallwachs discovered that when ultraviolet light (or any light of sufficient frequency) is directed at a metal surface, it causes the metal to emit electrons. This was the first evidence that light can cause electrons to be ejected from a metal.

Lenard’s Observation: Following Hallwachs, Philipp Lenard conducted experiments that confirmed Hallwachs' findings. Lenard observed that the emitted electrons (photoelectrons) had a range of energies and their number increased with the intensity of the incident light. He also found that the kinetic energy of these electrons depended on the frequency of the light, not its intensity.

Formula:

The relationship between the kinetic energy of the emitted electrons and the light is given by the photoelectric equation:

$ E_k = h\nu - \phi $

where:

$E_k $ is the kinetic energy of the emitted electrons,

h is Planck's constant,

$\nu $ is the frequency of the incident light,

$\phi$ is the work function of the metal.

These observations were fundamental in showing that light has particle-like properties, leading to the development of quantum mechanics.

11.4 Experimental Study of Photoelectric Effect: The photoelectric effect is a key experiment in understanding the dual nature of light. In this experiment, light is shone on a metal surface, and if the light has enough energy, it ejects electrons from the metal. 

1. Setup: A metal plate is exposed to light in a vacuum. The plate is connected to an electric circuit that measures the current of ejected electrons.

2. Observation: When light with a certain frequency hits the metal, electrons are emitted. The amount of current depends on the intensity of the light, while the frequency of light determines whether electrons are ejected at all.

3. Key Findings: The experiment shows that light behaves as particles called photons. Each photon has a specific amount of energy given by the formula:$ E = h\nu$

where E is the energy of the photon, h is Planck’s constant, and $\nu$ (nu) is the frequency of the light.

4. Photoelectric Equation: The kinetic energy of the emitted electrons can be calculated using:

$E_k = h\nu - \phi$

where $E_k$ is the kinetic energy of the ejected electron and $ \phi $ is the work function of the metal (the minimum energy required to eject an electron).

This experiment supports the idea that light behaves as particles, providing evidence for the dual nature of radiation and matter.

Effect of intensity of light on Photocurrent:

Higher Intensity: If the light’s intensity is increased, more photons hit the metal surface. This causes more electrons to be ejected, which increases the photocurrent.

Lower Intensity: If the light’s intensity is decreased, fewer photons hit the metal. This means fewer electrons are ejected, so the photocurrent is lower.

Formula: 

The photocurrent (I) is directly proportional to the light intensity ($I_0$):

$I \propto I_0 $

In short, more intense light produces a higher photocurrent because it ejects more electrons from the metal surface.

Effect of Potential on Photoelectric Current: It examines how changing the electric potential between the metal surface and the collector affects the current produced in the photoelectric effect.

Photoelectric Current: When light hits a metal surface, it releases electrons, creating a flow of current. This is called the photoelectric current.

Effect of Increasing Potential: If you increase the positive potential on the collector (relative to the metal surface), more emitted electrons are attracted to it, which increases the photoelectric current. 

Stopping Potential: If you increase the negative potential, it can stop the electrons from reaching the collector, reducing the current to zero. The potential needed to stop the current is called the stopping potential.

Formulas:

1. Photoelectric Current (I): It increases with the intensity of light and the positive potential applied.

2. Stopping Potential (V₀): This is related to the maximum kinetic energy of the emitted electrons:

$eV₀ = E_k $

where e is the electron charge and $ E_k $ is the maximum kinetic energy of the electrons.

In summary, the photoelectric current depends on the potential applied: increasing positive potential increases the current, while increasing negative potential can reduce it to zero.

Effect of Frequency of Incident Radiation on Stopping Potential: We study how the frequency of incident radiation (light) affects the stopping potential in the photoelectric effect. When light hits a metal surface, it can release electrons. The stopping potential is the voltage needed to stop these ejected electrons from reaching the anode.

Here’s how the frequency of the incident radiation affects the stopping potential:

1. Higher Frequency Increases Stopping Potential: As the frequency of the incident light increases, the energy of the photons increases. This means more energy is transferred to the electrons, making them move faster. Therefore, a higher stopping potential is required to stop these faster-moving electrons.

2. Formula to Understand the Relationship:

The energy of the incident photons is given by: $ E = h\nu $

where h is Planck’s constant and $ \nu$ is the frequency of the incident radiation.

The stopping potential $ V_s $ is related to the maximum kinetic energy of the emitted electrons: $ eV_s = h\nu - \phi $

where e is the charge of an electron, $ V_s$ is the stopping potential, and $ \phi $ is the work function of the metal.

11.5 Photoelectric Effect and Wave Theory of Light: 

Photoelectric Effect: The photoelectric effect occurs when light shines on a metal surface and causes the release of electrons from that surface. This phenomenon shows that light behaves like particles called photons. Each photon has a specific amount of energy.

Wave Theory of Light: According to the wave theory of light, light is considered as a wave that can carry energy. However, this theory struggled to explain why electrons are emitted only when the light frequency is above a certain level, regardless of the light's intensity.

Formula:

Energy of a Photon:$E = h\nu$

Here, E is the energy of the photon, h is Planck's constant, and $ \nu$ (nu) is the frequency of light.

Photoelectric Equation:

Kinetic Energy of Ejected Electrons: $ E_k = h\nu - \phi $

In this equation, $ E_k$ is the kinetic energy of the emitted electrons, $ \phi$ is the work function of the metal, and $h\nu $is the energy of the incoming photons.

The photoelectric effect provided evidence that light can behave like particles, challenging the old wave theory and leading to the development of quantum physics.

11.6 Einstein’s Photoelectric Equation: Energy Quantum of Radiation: Einstein’s photoelectric equation helps us understand how light can release electrons from a metal surface. According to Einstein, light is made of particles called photons, and each photon carries a specific amount of energy.

When light hits a metal, the energy from the photons can knock electrons off the metal surface. The energy of the photon must be greater than a certain threshold, known as the work function of the metal, for this to happen.

Formula: $ E_k = h\nu - \phi $

Here’s what each term means:

$ E_k$ is the kinetic energy of the ejected electron.

h is Planck’s constant.

$\nu $(nu) is the frequency of the incident light.

$ \phi $(phi) is the work function of the metal.

This equation shows that the energy of the incoming photons (light) is used to overcome the work function, and any extra energy is converted into the kinetic energy of the ejected electrons.

11.7 Particle Nature of Light: The Photon: The "Particle Nature of Light: The Photon" section explores the idea that light behaves as discrete packets of energy called photons. This concept helps explain how light interacts with matter in a particle-like manner.

Key Points:

Photon: A photon is a tiny packet of energy that makes up light. Each photon carries a specific amount of energy, which depends on the light's frequency.

Photoelectric Effect: When light hits a metal surface, it can eject electrons from the surface. This effect shows that light's energy comes in these packets or photons, not in a continuous wave. 

Energy of a Photon: The energy E  of a photon can be calculated using the formula:

$E = h\nu $

where h is Planck’s constant and $\nu $ (nu) is the frequency of the light.

This particle-like behaviour of light is crucial for understanding various physical phenomena and is a key part of the dual nature of radiation and matter.

Dual Nature of Matter and Radiation Class 12 Notes Physics - Basic Subjective Questions

Section-A (1 Mark Questions)

1. Define photoelectric work function. How is it related to threshold frequency?

Ans. The minimum amount of radiant energy needed to pull an electron (without imparting it any kinetic energy) from a metallic surface is called work function of the metal. The relation between work function W0 and threshold frequency ν0 is W0 = hν0. 

2. Which photon has more energy: A red one or a violet one? 

Ans. Violet photon has more energy, because energy of a photon, E = hν and νviolet > νred .  

3. Why are alkali metals most suited as photo-sensitive metals?

Ans. Alkali metals have low work function. Even visible radiation can eject out electrons from them. So, alkali metals are most suitable photo-sensitive metals. 

4. Work function of aluminium is 4.2 eV. If two photons each of energy 2.5 eV are incident on its surface, will the emission of electrons take place? Justify your answer.

Ans. No, energy of a single photon must be greater than the work function of the metal for the emission of a photoelectron. 

5. Does the 'stopping potential' in photo-electric emission depend upon (i) the intensity of the incident radiation in a photocell? (ii) the frequency of the incident radiation?

Ans. 

(i) No, the stopping potential does not depend on the intensity of the incident radiation. 

(ii) Yes, the stopping potential depends on the frequency of incident radiation, above the threshold frequency, V0 ∝ ν. 

6. How does the maximum kinetic energy of electrons emitted vary with the work function of the metal?

Ans. The maximum kinetic energy of emitted electrons, $K_{max}=\dfrac{1}{2}mv^{2}_{max}=hv-W_{0}$ .

7. The stopping potential in an experiment on photoelectric effect is 1.5 V. What is the maximum kinetic energy of the photoelectrons emitted?

Ans. $K_{max}=eV_{0}=1\cdot 6\times 10^{-19}C\times 1\cdot 5V$

$=24\times 10^{-19}J$ or 1.5eV

8. Why are de-Broglie waves associated with a moving football not visible? 

Ans. Due to the large mass of a football(in comparison subatomic particles), the de-Broglie wavelength $\left ( \lambda =\dfrac{h}{mv} \right )$ associated with a moving football is very small, quite beyond measurement. Hence its wave nature is not visible.

9. The de-Broglie wavelength of a particle of kinetic energy K is $\lambda$ . What would be the wavelength of the particle if its kinetic energy were K/4?

Ans. The de-Broglie wavelength of a particle of mass m and kinetic energy K is $\lambda=\dfrac{h}{\sqrt{2mK}}$ .

When the kinetic energy is $\dfrac{K}{4}$ .

${\lambda }'=\dfrac{h}{\sqrt{2m\dfrac{K}{4}}}=\dfrac{2h}{\sqrt{2mK}}=2\lambda$

10. An $\alpha -$ particle and a proton are accelerated through the same potential difference. Calculate the ratio of linear momenta acquired by the two. 

Ans. The kinetic energy gained by a particle when accelerated through potential difference V is $\dfrac{1}{2}mv^{2}=qV$ or $m^{2}v^{2}=2mqV$

∴ momentum $p=mv=\sqrt{2mqV}$

∴ $\dfrac{p\alpha }{p_{p}}=\sqrt{\dfrac{2m_{\alpha }q_{\alpha }V}{2m_{p}q_{p}V}}=\sqrt{\dfrac{2\times 4m_{p}\times 2e\times V}{2\times M_{P}\times e\times V}}=2\sqrt{2:1}$

Section-B (2 Marks Questions)

11. How is the photoelectric current affected on increasing the (i) frequency (ii) intensity of the incident radiations and why?

Ans.

(i) The increase in frequency of incident radiation has no effect on the photoelectric current. This is because the incident photon of increased energy cannot eject more than one electron from the metal surface. 

(ii) The photoelectric current increases proportionally with the increase in intensity of incident radiation. Larger the intensity of incident radiation, larger is the number of incident photons and hence larger is the number of electrons ejected from the metallic surface. 

12. What will happen to:

(i) kinetic energy of photoelectrons

(ii) photocurrent

If the light is changed from ultraviolet to X-rays in a photo-cell experiment? Intensity of the beam is the same in both cases.

Ans.

(i)As kinetic energy is proportional to the frequency of incident radiation and frequency of X-rays is higher than that of ultraviolet rays, so kinetic energy of photoelectrons will increase.

(ii) Photoelectric current will remain same as the intensity in both cases is same.

13. Two lines A and B in the plot given below show the variation of de-Broglie wavelength, $\lambda$ versus $\dfrac{1}{\sqrt{V}}$ where V is the accelerating potential difference, for two particles carrying the same charge. Which one of two represents a particle of smaller mass? 

Ans. According to de Broglie wavelength, $lambda =\dfrac{h}{\sqrt{2meV}}$

$\dfrac{\lambda }{1/\sqrt{V}}=\dfrac{h}{\sqrt{2me}}\Rightarrow \dfrac{\lambda }{1/\sqrt{V}}=\dfrac{1}{\sqrt{m}}\times \dfrac{h}{\sqrt{2e}}$

The slope is given by, $\dfrac{\lambda }{1/\sqrt{V}}=$ slope $\varpropto\dfrac{1}{\sqrt{m}}$

Slope of B > slope of A

14. The given graph shows the variation of photoelectric current (i) versus applied voltage (V) for two different photosensitive materials and for two different intensities of the incident radiation. Identify the pairs of curves that correspond to different materials but same intensity of incident radiation. 

Ans. The pairs (2, 4) and (1, 3) have same intensity but different material.  

15. Find the ratio of de-Broglie wavelengths associated with two electrons accelerated through 25 V and 36 V.

Ans. $\lambda \varpropto\dfrac{1}{\sqrt{V}}\dfrac{\lambda _{1}}{\lambda _{2}}=\sqrt{\dfrac{V_{2}}{V_{1}}}=\sqrt{\dfrac{36}{25}}=\dfrac{6}{5}$

$\therefore \lambda _{1}:\lambda _{2}::6:5$

Solved Sample NCERT Chapter 11 Questions Answers

Q1: The work function of a Cs (Cesium) metal is 2.14 eV. Light of frequency 6 x 1014 Hz incidents on it because of which emission of photoelectrons occurs. Find the following:

1. Maximum KE of the emitted electrons,

2. Stopping Potential, and

3. The maximum speed of emitted photoelectrons

Given: Here, Φo = 2.14 eV = 2.14 x 1.6 x 10-19J, 

f = 6 x 1014 Hz,

h (Planck’s Constant) = 6.6 x 10-34Js,

mass of an electron, me=  9.1 x 10-31 Kg

Sol:  

1.  Maximum KE =  hf - Φo = 

                              = 6.6 x 10-34Js x 6 x 1014 - 2.14 x 1.6 x 10-19J

                   Maximum KE   = 5.36 x 10-20J

 2. Stopping Potential Vo =\[\frac{K_{max}}{e}\]= (5.36 x 10-20/1.6 x 10-19)

                          Vo  = 0.335 eV

3. Maximum speed of emitted photoelectrons 

     We know that KE = ½ mv(max)2= 5.36 x 10-20J

    So, v(max)2 = (2 x 5.36 x 10-20/9.1 x 10-31)= 1.178 x 1011

or, vmax = 343 km/s

Q2: If the cut-off potential of a certain experiment is 1.5 V, then what will be the maximum KE of emitted photoelectrons?

Given:  Vo  =  1.5 V, e = 1.6 x 10-19C

To find: KEMAX =?

Solution: We know that KEMAX = e Vo 

                                                 = 1.6 x 10-19 x  1.5 

                                        KEMAX = 2.4 x 10-19J 

5 Important Formulas of Physics Class 12 Chapter 11 Dual Nature of Radiation and Matter

Importance of Chapter 11 Class 12 Physics Dual Nature of Radiation and Matter Notes PDF

• Revision notes help us quickly understand and remember key concepts before exams.

• They save time by focusing on essential information and skipping unnecessary details.

• They provide practical examples that show how theoretical knowledge is used in real-life situations.

• Revision notes ensure thorough preparation by covering all important topics in a structured manner.

• They increase confidence by clearly understanding what to expect in exams.

• Accessible formats like PDFs allow for easy studying anytime and anywhere.

Tips for Learning the Class 12 Physics Chapter 11 Notes on Dual Nature of Radiation and Matter

• Understand fundamental ideas like wave-particle duality, the photoelectric effect, and de Broglie’s hypothesis. Make sure you know how they explain the behaviour of light and matter.

• Learn and practise the key formulas related to this chapter, such as those for photon energy and de Broglie wavelength. Use flashcards or practice problems to keep them fresh in your memory.

• Study photoelectric effect and Compton scattering experiments. Understanding these experiments helps illustrate the dual nature of light and matter.

• Learn the concept that light and electrons exhibit both wave-like and particle-like properties. Make sure you can explain this duality in different contexts.

• Connect theoretical concepts to real-life applications and experiments. This makes the material more relatable and easier to remember.

Conclusion

Chapter 11 of Class 12 Physics Dual Nature of Radiation and Matter reveals the fascinating idea that light and matter exhibit both wave-like and particle-like properties. Our revision notes have covered essential topics such as the photoelectric effect, de Broglie’s wavelength, and the wave-particle duality, which are crucial for understanding modern physics. By mastering these concepts and formulas, you'll be well-prepared to tackle related problems and questions. Revisiting the key experiments and their implications will also deepen your learning of how these principles apply in real-world scenarios.

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