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Dual Nature of Radiation and Matter NCERT Solutions
If you are finding Class 12 Physics Chapter 11 a bit tricky with its shift from classical wave physics to modern quantum physics, don't worry at all. This comprehensive guide provides the complete, updated NCERT solutions, key formulas, conceptual breakdowns, and board-style extra questions. Perfect for your CBSE Board Exam 2026 and competitive exams like JEE and NEET. Let's simplify these deep modern physics concepts step-by-step-no blind rote memorization required. Toh chaliye, padhai shuru karein!
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Chapter NameDual Nature of Radiation and Matter (Chapter 11)
SubjectPhysics
ClassClass 12
BoardCBSE (Latest 2026-27 Syllabus)
Important TopicsPhotoelectric Effect, Einstein's Equation, de Broglie Wavelength
Difficulty LevelModerate (Highly conceptual with direct formula application)
Exam Weightage~5-6 Marks in Section G (Modern Physics Block)
Learning Objectives
After completing this chapter, students will be able to:
- Explain electron emission processes (Work Function, Thermionic, Field, and Photoelectric emission).
- Describe the experimental setup and observations of the photoelectric effect.
- Understand the threshold frequency, stopping potential, and its dependency on intensity and frequency.
- Explain why wave theory fails to explain the photoelectric effect and how Einstein's photon theory solves it.
- Calculate photon parameters like energy, momentum, and mass equivalents.
- Derive and apply the de Broglie equation to calculate the wavelength of moving matter particles (electrons, protons, etc.).
Key Concepts, Definitions & Formulas
- Work Function ($\phi_0$ or $W_0$): The minimum energy required by an electron to escape from a metal surface. It is measured in electron-volts ($\text{eV}$). Note that $1\text{eV} = 1.6 \times 10^{-19}\text{ J}$. Cesium has the lowest work function ($\sim 2.14\text{ eV}$), making it highly photosensitive.
- Photoelectric Effect: The phenomenon of emission of electrons from a clean metal surface when light of a sufficiently high and appropriate frequency falls on it.
- Threshold Frequency ($\nu_0$): The minimum frequency of incident radiation below which no photoelectric emission can occur, no matter how high the intensity is.
- Stopping Potential ($V_0$): The minimum negative (retarding) potential applied to the collector plate at which the photoelectric current drops to exactly zero. It measures the maximum kinetic energy of the emitted photoelectrons:
$$K_{\max} = eV_0$$ - Einstein's Photoelectric Equation:
$$h\nu = \phi_0 + K_{\max}$$
$$eV_0 = h\nu - h\nu_0$$ - Characteristics of Photons:
Energy of a photon: $E = h\nu = \frac{hc}{\lambda}$
Momentum of a photon: $p = \frac{E}{c} = \frac{h}{\lambda}$ - de Broglie Hypothesis (Dual Nature of Matter): The wavelength associated with a moving particle of mass $m$ and velocity $v$ is called the de Broglie wavelength:
$$\lambda = \frac{h}{p} = \frac{h}{mv}$$
In terms of Kinetic Energy ($K$):
$$\lambda = \frac{h}{\sqrt{2mK}}$$
For an electron accelerated through a potential difference of $V$ volts:
$$\lambda = \frac{1.227}{\sqrt{V}}\text{ nm} = \frac{12.27}{\sqrt{V}}\text{ \AA}$$
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Full NCERT Solutions
Question 11.1: Find the: (a) maximum frequency, and (b) minimum wavelength of X-rays produced by 30 kV electrons.
Answer:
Step 1: Write down the given values.
Accelerating potential, $V = 30\text{ kV} = 30 \times 10^3\text{ V}$
Electronic charge, $e = 1.6 \times 10^{-19}\text{ C}$
Planck's constant, $h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}$
Speed of light, $c = 3 \times 10^8\text{ m/s}$
The maximum kinetic energy of electrons striking the target is given by:
$$E_{\max} = eV = 1.6 \times 10^{-19} \times 30 \times 10^3 = 4.8 \times 10^{-15}\text{ J}$$
Step 2: Solve part (a) for Maximum Frequency ($\nu_{\max}$).
All the kinetic energy of the electron is converted into a single X-ray photon in the maximum frequency limit:
$$E_{\max} = h\nu_{\max} \implies \nu_{\max} = \frac{E_{\max}}{h}$$
$$\nu_{\max} = \frac{4.8 \times 10^{-15}}{6.63 \times 10^{-34}} \approx 7.24 \times 10^{18}\text{ Hz}$$
Step 3: Solve part (b) for Minimum Wavelength ($\lambda_{\min}$).
Using the relationship between wavelength, frequency, and the speed of light:
$$\lambda_{\min} = \frac{c}{\nu_{\max}}$$
$$\lambda_{\min} = \frac{3 \times 10^8}{7.24 \times 10^{18}} \approx 4.14 \times 10^{-11}\text{ m} = 0.0414\text{ nm}$$
Step 1: Write down the given values.
Accelerating potential, $V = 30\text{ kV} = 30 \times 10^3\text{ V}$
Electronic charge, $e = 1.6 \times 10^{-19}\text{ C}$
Planck's constant, $h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}$
Speed of light, $c = 3 \times 10^8\text{ m/s}$
The maximum kinetic energy of electrons striking the target is given by:
$$E_{\max} = eV = 1.6 \times 10^{-19} \times 30 \times 10^3 = 4.8 \times 10^{-15}\text{ J}$$
Step 2: Solve part (a) for Maximum Frequency ($\nu_{\max}$).
All the kinetic energy of the electron is converted into a single X-ray photon in the maximum frequency limit:
$$E_{\max} = h\nu_{\max} \implies \nu_{\max} = \frac{E_{\max}}{h}$$
$$\nu_{\max} = \frac{4.8 \times 10^{-15}}{6.63 \times 10^{-34}} \approx 7.24 \times 10^{18}\text{ Hz}$$
Step 3: Solve part (b) for Minimum Wavelength ($\lambda_{\min}$).
Using the relationship between wavelength, frequency, and the speed of light:
$$\lambda_{\min} = \frac{c}{\nu_{\max}}$$
$$\lambda_{\min} = \frac{3 \times 10^8}{7.24 \times 10^{18}} \approx 4.14 \times 10^{-11}\text{ m} = 0.0414\text{ nm}$$
Question 11.2: The work function of cesium metal is $2.14\text{ eV}$. When light of frequency $6 \times 10^{14}\text{ Hz}$ is incident on the metal surface, photoemission of electrons occurs. What is the: (a) maximum kinetic energy of the emitted electrons, (b) Stopping potential, and (c) maximum speed of the emitted photoelectrons?
Answer:
Step 1: Extract given parameters and convert units.
Work function, $\phi_0 = 2.14\text{ eV} = 2.14 \times 1.6 \times 10^{-19}\text{ J} = 3.424 \times 10^{-19}\text{ J}$
Frequency of incident light, $\nu = 6 \times 10^{14}\text{ Hz}$
Mass of an electron, $m = 9.1 \times 10^{-31}\text{ kg}$
Step 2: Calculate the energy of the incident photon ($E$).
$$E = h\nu = 6.63 \times 10^{-34} \times 6 \times 10^{14} = 3.978 \times 10^{-19}\text{ J}$$
Converting this energy into electron-volts ($\text{eV}$):
$$E = \frac{3.978 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.486\text{ eV}$$
Step 3: Solve part (a) for Maximum Kinetic Energy ($K_{\max}$).
According to Einstein's photoelectric equation:
$$K_{\max} = E - \phi_0$$
$$K_{\max} = 2.486\text{ eV} - 2.14\text{ eV} = 0.346\text{ eV}$$
In Joules:
$$K_{\max} = 0.346 \times 1.6 \times 10^{-19}\text{ J} \approx 5.54 \times 10^{-20}\text{ J}$$
Step 4: Solve part (b) for Stopping Potential ($V_0$).
We know that $K_{\max} = eV_0$.
$$0.346\text{ eV} = eV_0 \implies V_0 = 0.346\text{ V}$$
Step 5: Solve part (c) for Maximum Speed ($v_{\max}$).
Using the classical formula for kinetic energy:
$$K_{\max} = \frac{1}{2}mv_{\max}^2 \implies v_{\max} = \sqrt{\frac{2K_{\max}}{m}}$$
$$v_{\max} = \sqrt{\frac{2 \times 5.54 \times 10^{-20}}{9.1 \times 10^{-31}}} = \sqrt{12.18 \times 10^{10}} \approx 3.49 \times 10^5\text{ m/s}$$
Step 1: Extract given parameters and convert units.
Work function, $\phi_0 = 2.14\text{ eV} = 2.14 \times 1.6 \times 10^{-19}\text{ J} = 3.424 \times 10^{-19}\text{ J}$
Frequency of incident light, $\nu = 6 \times 10^{14}\text{ Hz}$
Mass of an electron, $m = 9.1 \times 10^{-31}\text{ kg}$
Step 2: Calculate the energy of the incident photon ($E$).
$$E = h\nu = 6.63 \times 10^{-34} \times 6 \times 10^{14} = 3.978 \times 10^{-19}\text{ J}$$
Converting this energy into electron-volts ($\text{eV}$):
$$E = \frac{3.978 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.486\text{ eV}$$
Step 3: Solve part (a) for Maximum Kinetic Energy ($K_{\max}$).
According to Einstein's photoelectric equation:
$$K_{\max} = E - \phi_0$$
$$K_{\max} = 2.486\text{ eV} - 2.14\text{ eV} = 0.346\text{ eV}$$
In Joules:
$$K_{\max} = 0.346 \times 1.6 \times 10^{-19}\text{ J} \approx 5.54 \times 10^{-20}\text{ J}$$
Step 4: Solve part (b) for Stopping Potential ($V_0$).
We know that $K_{\max} = eV_0$.
$$0.346\text{ eV} = eV_0 \implies V_0 = 0.346\text{ V}$$
Step 5: Solve part (c) for Maximum Speed ($v_{\max}$).
Using the classical formula for kinetic energy:
$$K_{\max} = \frac{1}{2}mv_{\max}^2 \implies v_{\max} = \sqrt{\frac{2K_{\max}}{m}}$$
$$v_{\max} = \sqrt{\frac{2 \times 5.54 \times 10^{-20}}{9.1 \times 10^{-31}}} = \sqrt{12.18 \times 10^{10}} \approx 3.49 \times 10^5\text{ m/s}$$
Question 11.3: The photoelectric threshold frequency of a certain metal is $3.3 \times 10^{14}\text{ Hz}$. If light of frequency $8.2 \times 10^{14}\text{ Hz}$ is incident on the metal, predict the cutoff voltage for the photoelectric emission.
Answer:
Step 1: Formulate expressions using given data.
Threshold frequency, $\nu_0 = 3.3 \times 10^{14}\text{ Hz}$
Incident frequency, $\nu = 8.2 \times 10^{14}\text{ Hz}$
Step 2: Set up Einstein's relation for cutoff/stopping voltage ($V_0$).
$$eV_0 = h(\nu - \nu_0) \implies V_0 = \frac{h(\nu - \nu_0)}{e}$$
Step 3: Calculate the value directly.
$$V_0 = \frac{6.63 \times 10^{-34} \times (8.2 \times 10^{14} - 3.3 \times 10^{14})}{1.6 \times 10^{-19}}$$
$$V_0 = \frac{6.63 \times 10^{-34} \times (4.9 \times 10^{14})}{1.6 \times 10^{-19}}$$
$$V_0 = \frac{3.2487 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.03\text{ V}$$
Step 1: Formulate expressions using given data.
Threshold frequency, $\nu_0 = 3.3 \times 10^{14}\text{ Hz}$
Incident frequency, $\nu = 8.2 \times 10^{14}\text{ Hz}$
Step 2: Set up Einstein's relation for cutoff/stopping voltage ($V_0$).
$$eV_0 = h(\nu - \nu_0) \implies V_0 = \frac{h(\nu - \nu_0)}{e}$$
Step 3: Calculate the value directly.
$$V_0 = \frac{6.63 \times 10^{-34} \times (8.2 \times 10^{14} - 3.3 \times 10^{14})}{1.6 \times 10^{-19}}$$
$$V_0 = \frac{6.63 \times 10^{-34} \times (4.9 \times 10^{14})}{1.6 \times 10^{-19}}$$
$$V_0 = \frac{3.2487 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.03\text{ V}$$
Question 11.4: The work function of a certain metal is $4.2\text{ eV}$. Will this metal give photoelectric emission for incident radiation of wavelength $330\text{ nm}$?
Answer:
Step 1: Write given values and target conversion parameters.
Work function, $\phi_0 = 4.2\text{ eV}$
Incident wavelength, $\lambda = 330\text{ nm} = 330 \times 10^{-9}\text{ m}$
Step 2: Calculate the energy of the incident photon ($E$).
$$E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{330 \times 10^{-9}}$$
$$E = \frac{1.989 \times 10^{-25}}{3.3 \times 10^{-7}} = 6.027 \times 10^{-19}\text{ J}$$
Step 3: Convert the photon energy to electron-volts.
$$E = \frac{6.027 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.77\text{ eV}$$
Step 4: Analyze the condition for photoemission.
Photoelectric emission can only occur if the energy of the incident photon is greater than or equal to the work function ($E \ge \phi_0$).
Here, $3.77\text{ eV} < 4.2\text{ eV}$.
Conclusion: No, the metal will not show any photoelectric emission for this radiation.
Step 1: Write given values and target conversion parameters.
Work function, $\phi_0 = 4.2\text{ eV}$
Incident wavelength, $\lambda = 330\text{ nm} = 330 \times 10^{-9}\text{ m}$
Step 2: Calculate the energy of the incident photon ($E$).
$$E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{330 \times 10^{-9}}$$
$$E = \frac{1.989 \times 10^{-25}}{3.3 \times 10^{-7}} = 6.027 \times 10^{-19}\text{ J}$$
Step 3: Convert the photon energy to electron-volts.
$$E = \frac{6.027 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.77\text{ eV}$$
Step 4: Analyze the condition for photoemission.
Photoelectric emission can only occur if the energy of the incident photon is greater than or equal to the work function ($E \ge \phi_0$).
Here, $3.77\text{ eV} < 4.2\text{ eV}$.
Conclusion: No, the metal will not show any photoelectric emission for this radiation.
Question 11.5: Light of frequency $7.21 \times 10^{14}\text{ Hz}$ is incident on a metal surface. Electrons with a maximum speed of $6.0 \times 10^5\text{ m/s}$ are ejected from the surface. What is the threshold frequency for photoemission of electrons?
Answer:
Step 1: Gather parameters.
Incident frequency, $\nu = 7.21 \times 10^{14}\text{ Hz}$
Maximum velocity, $v_{\max} = 6.0 \times 10^5\text{ m/s}$
Electron mass, $m = 9.1 \times 10^{-31}\text{ kg}$
Step 2: Compute Maximum Kinetic Energy ($K_{\max}$).
$$K_{\max} = \frac{1}{2}mv_{\max}^2 = \frac{1}{2} \times (9.1 \times 10^{-31}) \times (6.0 \times 10^5)^2$$
$$K_{\max} = 0.5 \times 9.1 \times 10^{-31} \times 3.6 \times 10^{11} = 1.638 \times 10^{-19}\text{ J}$$
Step 3: Convert kinetic energy into frequency equivalents.
According to Einstein's equation: $h\nu = h\nu_0 + K_{\max} \implies h\nu_0 = h\nu - K_{\max}$
$$\nu_0 = \nu - \frac{K_{\max}}{h}$$
$$\nu_0 = 7.21 \times 10^{14} - \frac{1.638 \times 10^{-19}}{6.63 \times 10^{-34}}$$
$$\nu_0 = 7.21 \times 10^{14} - 0.247 \times 10^{15} = 7.21 \times 10^{14} - 2.47 \times 10^{14}$$
$$\nu_0 = 4.74 \times 10^{14}\text{ Hz}$$
Step 1: Gather parameters.
Incident frequency, $\nu = 7.21 \times 10^{14}\text{ Hz}$
Maximum velocity, $v_{\max} = 6.0 \times 10^5\text{ m/s}$
Electron mass, $m = 9.1 \times 10^{-31}\text{ kg}$
Step 2: Compute Maximum Kinetic Energy ($K_{\max}$).
$$K_{\max} = \frac{1}{2}mv_{\max}^2 = \frac{1}{2} \times (9.1 \times 10^{-31}) \times (6.0 \times 10^5)^2$$
$$K_{\max} = 0.5 \times 9.1 \times 10^{-31} \times 3.6 \times 10^{11} = 1.638 \times 10^{-19}\text{ J}$$
Step 3: Convert kinetic energy into frequency equivalents.
According to Einstein's equation: $h\nu = h\nu_0 + K_{\max} \implies h\nu_0 = h\nu - K_{\max}$
$$\nu_0 = \nu - \frac{K_{\max}}{h}$$
$$\nu_0 = 7.21 \times 10^{14} - \frac{1.638 \times 10^{-19}}{6.63 \times 10^{-34}}$$
$$\nu_0 = 7.21 \times 10^{14} - 0.247 \times 10^{15} = 7.21 \times 10^{14} - 2.47 \times 10^{14}$$
$$\nu_0 = 4.74 \times 10^{14}\text{ Hz}$$
Question 11.6: Light of wavelength $488\text{ nm}$ is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is $0.38\text{ V}$. Find the work function of the material from which the emitter is made.
Answer:
Step 1: Set up given parameters.
Wavelength, $\lambda = 488\text{ nm} = 488 \times 10^{-9}\text{ m}$
Stopping potential, $V_0 = 0.38\text{ V} \implies K_{\max} = 0.38\text{ eV}$
Step 2: Calculate the incident photon energy ($E$).
$$E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{488 \times 10^{-9}} = 4.075 \times 10^{-19}\text{ J}$$
Converting to $\text{eV}$:
$$E = \frac{4.075 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.55\text{ eV}$$
Step 3: Find work function ($\phi_0$).
$$E = \phi_0 + K_{\max} \implies \phi_0 = E - K_{\max}$$
$$\phi_0 = 2.55\text{ eV} - 0.38\text{ eV} = 2.17\text{ eV}$$
Step 1: Set up given parameters.
Wavelength, $\lambda = 488\text{ nm} = 488 \times 10^{-9}\text{ m}$
Stopping potential, $V_0 = 0.38\text{ V} \implies K_{\max} = 0.38\text{ eV}$
Step 2: Calculate the incident photon energy ($E$).
$$E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{488 \times 10^{-9}} = 4.075 \times 10^{-19}\text{ J}$$
Converting to $\text{eV}$:
$$E = \frac{4.075 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.55\text{ eV}$$
Step 3: Find work function ($\phi_0$).
$$E = \phi_0 + K_{\max} \implies \phi_0 = E - K_{\max}$$
$$\phi_0 = 2.55\text{ eV} - 0.38\text{ eV} = 2.17\text{ eV}$$
Question 11.7: What is the de Broglie wavelength of: (a) a bullet of mass $0.040\text{ kg}$ traveling at the speed of $1.0\text{ km/s}$, (b) a ball of mass $0.060\text{ kg}$ moving at a speed of $1.0\text{ m/s}$, and (c) a dust particle of mass $1.0 \times 10^{-9}\text{ kg}$ drifting with a speed of $2.2\text{ m/s}$?
Answer:
Using de Broglie's relation: $\lambda = \frac{h}{mv}$
Step 1: Solve part (a) for the bullet.
$m = 0.040\text{ kg}$, $v = 1.0\text{ km/s} = 1000\text{ m/s}$
$$\lambda = \frac{6.63 \times 10^{-34}}{0.040 \times 1000} = \frac{6.63 \times 10^{-34}}{40} \approx 1.66 \times 10^{-35}\text{ m}$$
Step 2: Solve part (b) for the ball.
$m = 0.060\text{ kg}$, $v = 1.0\text{ m/s}$
$$\lambda = \frac{6.63 \times 10^{-34}}{0.060 \times 1.0} \approx 1.11 \times 10^{-32}\text{ m}$$
Step 3: Solve part (c) for the dust particle.
$m = 1.0 \times 10^{-9}\text{ kg}$, $v = 2.2\text{ m/s}$
$$\lambda = \frac{6.63 \times 10^{-34}}{1.0 \times 10^{-9} \times 2.2} \approx 3.01 \times 10^{-25}\text{ m}$$
(Note: These wavelengths are immensely small and completely unmeasurable, explaining why macroscopic objects do not display prominent wave characteristics in daily life).
Using de Broglie's relation: $\lambda = \frac{h}{mv}$
Step 1: Solve part (a) for the bullet.
$m = 0.040\text{ kg}$, $v = 1.0\text{ km/s} = 1000\text{ m/s}$
$$\lambda = \frac{6.63 \times 10^{-34}}{0.040 \times 1000} = \frac{6.63 \times 10^{-34}}{40} \approx 1.66 \times 10^{-35}\text{ m}$$
Step 2: Solve part (b) for the ball.
$m = 0.060\text{ kg}$, $v = 1.0\text{ m/s}$
$$\lambda = \frac{6.63 \times 10^{-34}}{0.060 \times 1.0} \approx 1.11 \times 10^{-32}\text{ m}$$
Step 3: Solve part (c) for the dust particle.
$m = 1.0 \times 10^{-9}\text{ kg}$, $v = 2.2\text{ m/s}$
$$\lambda = \frac{6.63 \times 10^{-34}}{1.0 \times 10^{-9} \times 2.2} \approx 3.01 \times 10^{-25}\text{ m}$$
(Note: These wavelengths are immensely small and completely unmeasurable, explaining why macroscopic objects do not display prominent wave characteristics in daily life).
Question 11.8: An electron, an $\alpha$-particle, and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength?
Answer:
Step 1: Establish the mathematical proportional dependency.
The formula linking de Broglie wavelength with kinetic energy ($K$) is:
$$\lambda = \frac{h}{\sqrt{2mK}}$$
Step 2: Compare parameters.
Since the kinetic energy ($K$) and Planck's constant ($h$) are completely identical for all three particles:
$$\lambda \propto \frac{1}{\sqrt{m}}$$
Step 3: Rank by masses.
Let's look at their masses:
* Mass of electron ($m_e$) is the smallest.
* Mass of proton ($m_p$) is intermediate.
* Mass of alpha particle ($m_\alpha \approx 4m_p$) is the largest.
Since mass is inversely proportional to wavelength, the particle with the largest mass will have the shortest wavelength.
Conclusion: The $\alpha$-particle has the largest mass and therefore features the shortest de Broglie wavelength.
Step 1: Establish the mathematical proportional dependency.
The formula linking de Broglie wavelength with kinetic energy ($K$) is:
$$\lambda = \frac{h}{\sqrt{2mK}}$$
Step 2: Compare parameters.
Since the kinetic energy ($K$) and Planck's constant ($h$) are completely identical for all three particles:
$$\lambda \propto \frac{1}{\sqrt{m}}$$
Step 3: Rank by masses.
Let's look at their masses:
* Mass of electron ($m_e$) is the smallest.
* Mass of proton ($m_p$) is intermediate.
* Mass of alpha particle ($m_\alpha \approx 4m_p$) is the largest.
Since mass is inversely proportional to wavelength, the particle with the largest mass will have the shortest wavelength.
Conclusion: The $\alpha$-particle has the largest mass and therefore features the shortest de Broglie wavelength.
Question 11.9: An electron is accelerated through a potential difference of $100\text{ V}$. What is its de Broglie wavelength?
Answer:
Step 1: Identify acceleration parameter.
Potential difference, $V = 100\text{ V}$
Step 2: Apply the simplified specific wavelength formula for accelerated electrons.
$$\lambda = \frac{1.227}{\sqrt{V}}\text{ nm}$$
Step 3: Compute the square root and find the answer.
$$\lambda = \frac{1.227}{\sqrt{100}} = \frac{1.227}{10} = 0.1227\text{ nm} = 1.227\text{ \AA}$$
Step 1: Identify acceleration parameter.
Potential difference, $V = 100\text{ V}$
Step 2: Apply the simplified specific wavelength formula for accelerated electrons.
$$\lambda = \frac{1.227}{\sqrt{V}}\text{ nm}$$
Step 3: Compute the square root and find the answer.
$$\lambda = \frac{1.227}{\sqrt{100}} = \frac{1.227}{10} = 0.1227\text{ nm} = 1.227\text{ \AA}$$
Extra Important Questions (Board Style 2026)
Section A: Multiple Choice Questions (MCQs)
1. When the intensity of an incident thin monochromatic light beam hitting a photosensitive emitter is doubled, the maximum kinetic energy of the photoelectrons will:
A) Double
B) Quadruple
C) Remain completely unchanged
D) Become halved
Answer: C
Explanation: Maximum kinetic energy depends purely on the frequency of the incident light and work function of the plate, not on its intensity. Intensity only changes the number of emitted photons, thus doubling the current. (Difficulty Level: Easy)
Explanation: Maximum kinetic energy depends purely on the frequency of the incident light and work function of the plate, not on its intensity. Intensity only changes the number of emitted photons, thus doubling the current. (Difficulty Level: Easy)
2. The stopping potential ($V_0$) versus frequency ($\nu$) plot for a photosensitive metal gives a straight line graph. The slope of this line represents:
A) $\phi_0$
B) $h/e$
C) $e/h$
D) $h$
Answer: B
Explanation: From Einstein's equation, $V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e}$. Comparing this with the slope-intercept form $y = mx + c$, the slope $m = h/e$. (Difficulty Level: Medium)
Explanation: From Einstein's equation, $V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e}$. Comparing this with the slope-intercept form $y = mx + c$, the slope $m = h/e$. (Difficulty Level: Medium)
3. If a proton and an electron have identical de Broglie wavelengths, the ratio of their momentum values ($p_p/p_e$) is:
A) 1
B) 1836
C) 1/1836
D) Dependent on accelerating configurations
Answer: A
Explanation: Since $\lambda = h/p$, matching wavelengths instantly implies completely matching linear momentum magnitudes ($p = h/\lambda$). (Difficulty Level: Easy)
Explanation: Since $\lambda = h/p$, matching wavelengths instantly implies completely matching linear momentum magnitudes ($p = h/\lambda$). (Difficulty Level: Easy)
Section B: Short Answer Questions
4. Why is the wave theory of light completely unable to explain the existence of threshold frequency during photoelectric emissions?
Answer:
Step 1: Classical View. According to classical wave theory, energy carried by a wave is proportional to its intensity. Thus, a high-intensity beam of low frequency should ideally accumulate enough energy over time to eject an electron.
Step 2: Experimental Reality. However, experiments show that if the frequency is below the threshold value, no emission occurs regardless of intensity. Wave theory fails because it expects continuous energy absorption instead of a discrete particle-like quantum collision.
Step 1: Classical View. According to classical wave theory, energy carried by a wave is proportional to its intensity. Thus, a high-intensity beam of low frequency should ideally accumulate enough energy over time to eject an electron.
Step 2: Experimental Reality. However, experiments show that if the frequency is below the threshold value, no emission occurs regardless of intensity. Wave theory fails because it expects continuous energy absorption instead of a discrete particle-like quantum collision.
5. Define the term 'Intensity of Light' according to quantum photon theory and state its unit.
Answer:
Step 1: Definition. According to photon theory, the intensity of light ($I$) is defined as the total number of photons crossing a unit area normally per unit time multiplied by the energy of each photon.
$$I = \frac{N \cdot h\nu}{A \cdot t}$$
Step 2: Unit. Its standard SI unit is Watts per square meter ($\text{W/m}^2$).
Step 1: Definition. According to photon theory, the intensity of light ($I$) is defined as the total number of photons crossing a unit area normally per unit time multiplied by the energy of each photon.
$$I = \frac{N \cdot h\nu}{A \cdot t}$$
Step 2: Unit. Its standard SI unit is Watts per square meter ($\text{W/m}^2$).
6. Calculate the de Broglie wavelength of a thermal neutron at room temperature ($27^\circ\text{C}$). [Given: $m_n = 1.67 \times 10^{-27}\text{ kg}$, $k_B = 1.38 \times 10^{-23}\text{ J/K}$]
Answer:
Step 1: Convert temperature to Kelvin. $T = 27 + 273 = 300\text{ K}$.
Step 2: Calculate Kinetic Energy. Kinetic energy of a thermal neutron is given by $K = \frac{3}{2}k_B T$.
Step 3: Use the de Broglie relation.
$$\lambda = \frac{h}{\sqrt{2m_n K}} = \frac{h}{\sqrt{3m_n k_B T}}$$
$$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{3 \times 1.67 \times 10^{-27} \times 1.38 \times 10^{-23} \times 300}} \approx 0.145\text{ nm}$$
Step 1: Convert temperature to Kelvin. $T = 27 + 273 = 300\text{ K}$.
Step 2: Calculate Kinetic Energy. Kinetic energy of a thermal neutron is given by $K = \frac{3}{2}k_B T$.
Step 3: Use the de Broglie relation.
$$\lambda = \frac{h}{\sqrt{2m_n K}} = \frac{h}{\sqrt{3m_n k_B T}}$$
$$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{3 \times 1.67 \times 10^{-27} \times 1.38 \times 10^{-23} \times 300}} \approx 0.145\text{ nm}$$
7. Work functions of two metals A and B are $2\text{ eV}$ and $4\text{ eV}$ respectively. Which metal has a lower threshold wavelength?
Answer:
Step 1: Relate work function to threshold wavelength. We know that $\phi_0 = \frac{hc}{\lambda_0}$, which implies $\lambda_0 \propto \frac{1}{\phi_0}$.
Step 2: Compare values. Therefore, a higher work function corresponds to a shorter threshold wavelength. Since metal B has a larger work function ($4\text{ eV} > 2\text{ eV}$), Metal B has the lower threshold wavelength.
Step 1: Relate work function to threshold wavelength. We know that $\phi_0 = \frac{hc}{\lambda_0}$, which implies $\lambda_0 \propto \frac{1}{\phi_0}$.
Step 2: Compare values. Therefore, a higher work function corresponds to a shorter threshold wavelength. Since metal B has a larger work function ($4\text{ eV} > 2\text{ eV}$), Metal B has the lower threshold wavelength.
Section C: Long Answer Questions
8. Draw a neat experimental layout diagram of the photoelectric effect. Plot graphs showing variation of photoelectric current with (i) Collector plate potential for different intensities, and (ii) Collector plate potential for different frequencies.
Answer:
Experimental Diagram Details: Show an evacuated glass tube containing a photosensitive emitter plate and a collector plate. Include a quartz window to let monochromatic light pass through, along with a voltmeter, ammeter, potential divider, and a commutator switch to reverse polarities.
Graph (i): Current vs Potential for varying intensities ($I_1 < I_2 < I_3$) at fixed frequency:
Graph (ii): Current vs Potential for varying frequencies ($\nu_1 < \nu_2 < \nu_3$) at fixed intensity:
Experimental Diagram Details: Show an evacuated glass tube containing a photosensitive emitter plate and a collector plate. Include a quartz window to let monochromatic light pass through, along with a voltmeter, ammeter, potential divider, and a commutator switch to reverse polarities.
Graph (i): Current vs Potential for varying intensities ($I_1 < I_2 < I_3$) at fixed frequency:
Photoelectric Current ^ | _________ I3 | /_________ I2 | //_________ I1 | /// ----|------///--------> Collector Potential (V) -V0| ///(Observation: Stopping potential $-V_0$ stays identical, but saturation current increases with intensity).
Graph (ii): Current vs Potential for varying frequencies ($\nu_1 < \nu_2 < \nu_3$) at fixed intensity:
Photoelectric Current ^ | Saturation Current | /// | /// ----|---|---|-----------> Collector Potential (V) -V03 -V02 -V01(Observation: Higher frequencies shift the stopping potentials more negative, while the saturation current remains constant).
9. State the three main features of the photoelectric effect that cannot be explained by wave theory. Briefly explain how Einstein's photon theory successfully accounts for them.
Answer:
The three features that wave theory fails to explain are:
1. Instantaneous Emission: Wave theory requires electrons to gradually absorb energy over time before escaping. In reality, emission happens instantly ($\sim 10^{-9}\text{ s}$), which photon theory explains as a one-on-one elastic collision between a photon and an electron.
2. Existence of Threshold Frequency: Wave theory suggests any frequency can cause emission if the intensity is high enough. Photon theory states that if a single photon's energy ($h\nu$) is less than the work function ($\phi_0$), no electron can escape.
3. Dependency of Maximum Kinetic Energy on Frequency: Wave theory predicts that kinetic energy should increase with intensity. In reality, it depends linearly on frequency, as shown by Einstein's equation $K_{\max} = h\nu - \phi_0$.
The three features that wave theory fails to explain are:
1. Instantaneous Emission: Wave theory requires electrons to gradually absorb energy over time before escaping. In reality, emission happens instantly ($\sim 10^{-9}\text{ s}$), which photon theory explains as a one-on-one elastic collision between a photon and an electron.
2. Existence of Threshold Frequency: Wave theory suggests any frequency can cause emission if the intensity is high enough. Photon theory states that if a single photon's energy ($h\nu$) is less than the work function ($\phi_0$), no electron can escape.
3. Dependency of Maximum Kinetic Energy on Frequency: Wave theory predicts that kinetic energy should increase with intensity. In reality, it depends linearly on frequency, as shown by Einstein's equation $K_{\max} = h\nu - \phi_0$.
Section D: Case-Based Questions
Case Study: Photovoltaic Sensors in Automated Infrastructure
Modern automated city infrastructure relies heavily on light-dependent sensors. Street lights automatically switch off at dawn using photocells. In these cells, an incident light beam strikes a potassium or cesium-coated surface inside an evacuated bulb, triggering an electric current that controls a relay switch.
Q10. Which physical parameter of the light determines whether the street light sensor triggers at dawn?
Answer: The frequency/wavelength of the solar radiation. It must exceed the threshold frequency of the photoemissive material.
Q11. If the street light sensor window is covered by a dirty, semi-translucent film that reduces light intensity by half without changing its color temperature, will the sensor still function?
Answer: Yes, it will. Reducing intensity decreases the overall current, but it does not change the energy of individual photons. As long as the frequency remains above the threshold value, photoelectrons will still be emitted to trigger the circuit.
Q12. Why are alkali metals preferred for constructing photocell surfaces?
Answer: Alkali metals like cesium and potassium have very low work functions. This allows them to emit electrons even when exposed to low-energy visible light, making them ideal for solar sensors.
Answer: The frequency/wavelength of the solar radiation. It must exceed the threshold frequency of the photoemissive material.
Q11. If the street light sensor window is covered by a dirty, semi-translucent film that reduces light intensity by half without changing its color temperature, will the sensor still function?
Answer: Yes, it will. Reducing intensity decreases the overall current, but it does not change the energy of individual photons. As long as the frequency remains above the threshold value, photoelectrons will still be emitted to trigger the circuit.
Q12. Why are alkali metals preferred for constructing photocell surfaces?
Answer: Alkali metals like cesium and potassium have very low work functions. This allows them to emit electrons even when exposed to low-energy visible light, making them ideal for solar sensors.
Section E: Assertion-Reason Questions
Directions: (A) Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A. (B) Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A. (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.
13. Assertion (A): Light possesses dual nature, displaying both wave behavior and particle behavior under different conditions. Reason (R): Phenomena like diffraction and interference demonstrate the particle-like behavior of light, whereas the photoelectric effect highlights its wave nature.
Answer: C
Explanation: The assertion is true. However, the reason is completely false because interference and diffraction demonstrate the wave nature of light, while the photoelectric effect demonstrates its particle nature.
Explanation: The assertion is true. However, the reason is completely false because interference and diffraction demonstrate the wave nature of light, while the photoelectric effect demonstrates its particle nature.
14. Assertion (A): Matter waves are not electromagnetic waves. Reason (R): Electromagnetic waves are produced by accelerated charges, whereas matter waves are associated with any moving particle regardless of its charge state.
Answer: A
Explanation: Matter waves travel with velocities different from the speed of light and do not require charge acceleration to exist, unlike EM waves. The reason correctly explains the assertion.
Explanation: Matter waves travel with velocities different from the speed of light and do not require charge acceleration to exist, unlike EM waves. The reason correctly explains the assertion.
15. Assertion (A): The de Broglie wavelength of an electron accelerated to a high potential varies inversely with the square root of the potential. Reason (R): Momentum of the accelerated particle is directly proportional to the applied voltage.
Answer: C
Explanation: The assertion is true ($\lambda \propto 1/\sqrt{V}$). The reason is false because momentum $p = \sqrt{2meV}$, meaning it is proportional to the square root of the voltage, not directly proportional.
Explanation: The assertion is true ($\lambda \propto 1/\sqrt{V}$). The reason is false because momentum $p = \sqrt{2meV}$, meaning it is proportional to the square root of the voltage, not directly proportional.
Common Mistakes Students Make
- Mixing up Units (eV and Joules): This is the most frequent calculation mistake. When using $E = h\nu$ or $K_{\max} = \frac{1}{2}mv^2$, your calculations are in Joules. Do not directly subtract a work function given in eV from an energy value calculated in Joules without converting it first! Always multiply eV by $1.6 \times 10^{-19}$ to get Joules.
- Confusing Intensity and Frequency: Remember this simple rule: Frequency controls the Speed/Energy of the electrons. Intensity controls the Number of electrons (current).
- Incorrect de Broglie Formulas: Students often use the electron formula ($\lambda = 1.227/\sqrt{V}\text{ nm}$) for other particles like protons or alpha particles. This simplified equation is calibrated only for an electron's mass and charge. For protons, use the full equation $\lambda = h / \sqrt{2mKV}$.
Exam Preparation Tips
- Master the Dual Graph Variations: Expect at least one graph-based question from the experimental section of this chapter. Practice drawing the current versus voltage curves precisely.
- Memorize Key Constants: Keep value approximations ready for quick calculations during your exam: $hc \approx 19.89 \times 10^{-26}\text{ J}\cdot\text{m}$. In eV-nanometers, $hc \approx 1240\text{ eV}\cdot\text{nm}$. This shortcut saves a lot of time during numerical calculations.
- Focus on Reasoning Questions: Modern physics sections feature fewer derivations and more conceptual reasoning questions. Focus on explaining why wave theory fails and how the photon model provides the correct explanation.
FAQ Section
Is Chapter 11 Dual Nature of Radiation and Matter important for CBSE Class 12 boards?
Yes, it is highly scoring. It serves as the foundation for the Modern Physics unit, and its numerical questions are direct and formula-based.
Where can I download the updated NCERT solutions PDF for Class 12 Physics?
You can easily save this page or print it directly using your browser's print options to use it as a complete study guide.
What is the physical significance of the de Broglie wavelength?
It shows that matter is not just composed of rigid particles; moving matter also has wave properties. This dual nature forms the foundation of modern quantum mechanics and technologies like electron microscopes.
Why do we not see wave properties in daily objects like a moving cricket ball?
Because macroscopic objects have a very large mass. Since $\lambda = h/mv$, a large mass makes the wavelength extremely small ($\sim 10^{-34}\text{ m}$), making their wave properties unnoticeable.
What is wattless current and does it relate to this chapter?
Wattless current occurs in AC circuits containing only inductors or capacitors where power loss is zero. It is an electrical circuit concept from Chapter 7, not related to the quantum mechanics of Chapter 11.
Conclusion: Mastering Chapter 11 Dual Nature of Radiation and Matter is all about understanding the balance between particle behavior and wave behavior. Make sure to practice the numerical exercises by hand and review past board questions. Save this guide for your revisions, stay confident, and good luck with your preparation!
More Class 12 Physics Chapters
Ch 1: Electric Charges and Fields
Ch 2: Electrostatic Potential and Capacitance
Ch 3: Current Electricity
Ch 4: Moving Charges and Magnetism
Ch 5: Magnetism and Matter
Ch 6: Electromagnetic Induction
Ch 7: Alternating Current
Ch 8: Electromagnetic Waves
Ch 9: Ray Optics
Ch 10: Wave Optics
Ch 11: Dual Nature of Radiation and Matter
Ch 12: Atoms
Ch 13: Nuclei
Class 12 Mock QUIZs
Ch 14: Semiconductor Electronics