Table of Contents
How to Use This Page
Read each concept carefully, then check the formula, common mistake, and exam tip before moving to the next. This page completely covers the Dual Nature of Radiation and Matter for CBSE Class 12 Physics, marking the birth of Quantum Mechanics.
Key Concepts
The Birth of Quantum Physics
When waves act like particles, and particles act like waves
1
Work Function ($\Phi_0$ or $W$)
Definition
The minimum amount of energy required by an electron to just escape from the metal surface. It depends on the nature of the metal and its surface conditions. Measured in electron-Volts ($\text{eV}$).
$$1 \text{ eV} = 1.6 \times 10^{-19} \text{ Joules}$$
2
The Photoelectric Effect
Concept
The phenomenon of emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequency (like UV light) falls on it. The emitted electrons are called photoelectrons.
$$\text{Light (Photons)} \rightarrow \text{Metal} \rightarrow \text{Electrons (Current)}$$
3
Threshold Frequency ($\nu_0$) & Wavelength ($\lambda_0$)
Formula
The minimum frequency ($\nu_0$) or maximum wavelength ($\lambda_0$) of incident light required to just eject an electron without giving it any kinetic energy.
$$\Phi_0 = h\nu_0 = \frac{hc}{\lambda_0}$$
4
Einstein’s Photoelectric Equation
Formula
Based on the conservation of energy. The energy of an incident photon ($h\nu$) is used in two ways: first, to overcome the work function ($\Phi_0$), and second, the remainder becomes the maximum kinetic energy of the electron.
$$K_{\text{max}} = h\nu – \Phi_0 = h(\nu – \nu_0)$$
5
Stopping Potential (Cut-off Voltage, $V_0$)
Formula
The minimum negative (retarding) potential applied to the anode with respect to the cathode that completely stops even the most energetic photoelectrons from reaching it. It is a direct measure of $K_{\text{max}}$.
$$K_{\text{max}} = e V_0 \quad \implies \quad e V_0 = h\nu – \Phi_0$$
6
Properties of Photons
Formula
Light consists of discrete energy packets called photons. They travel at the speed of light ($c$). Their rest mass is exactly zero, but they possess momentum.
$$E = h\nu = \frac{hc}{\lambda} \quad | \quad p = \frac{E}{c} = \frac{h}{\lambda}$$
7
De Broglie Wavelength of Matter Waves
Formula
Louis de Broglie proposed that if light waves can act like particles (photons), then moving material particles (like electrons) must act like waves.
$$\lambda = \frac{h}{p} = \frac{h}{mv} = \frac{h}{\sqrt{2mK}}$$
8
Wavelength of an Accelerated Electron
Formula
If an electron is accelerated from rest through a potential difference $V$, its kinetic energy becomes $K = eV$. Substituting constants ($h, m_e, e$) yields a powerful shortcut.
$$\lambda = \frac{h}{\sqrt{2meV}} \approx \frac{1.227}{\sqrt{V}} \text{ nm} \quad \text{or} \quad \frac{12.27}{\sqrt{V}} \text{ \AA}$$
Concept Deep Dive
01
Intensity vs. Frequency (The Classical Failure)
Why classical physics couldn’t explain itAccording to classical wave theory, a brighter (more intense) light has more energy. If you shine a super bright red light on a metal, the electrons should eventually absorb enough energy to pop out.
Experimental Reality: No matter how bright the red light is, or how long you wait, NO electrons are emitted if the frequency is below the threshold ($\nu_0$). But even the dimmest, weakest ultraviolet light pops electrons out instantly!
Einstein’s Genius: He realized Intensity just means “more photons per second,” while Frequency dictates the “energy of each individual photon.” If a single photon doesn’t have enough energy ($\Phi_0$) to break the electron free, throwing a billion of them at the metal (high intensity) still won’t work, because an electron can only absorb ONE photon at a time.
Experimental Reality: No matter how bright the red light is, or how long you wait, NO electrons are emitted if the frequency is below the threshold ($\nu_0$). But even the dimmest, weakest ultraviolet light pops electrons out instantly!
Einstein’s Genius: He realized Intensity just means “more photons per second,” while Frequency dictates the “energy of each individual photon.” If a single photon doesn’t have enough energy ($\Phi_0$) to break the electron free, throwing a billion of them at the metal (high intensity) still won’t work, because an electron can only absorb ONE photon at a time.
02
Reading the Stopping Potential Graph
Finding Planck’s Constant experimentallyIf you plot Stopping Potential ($V_0$) on the Y-axis and Frequency ($\nu$) on the X-axis, you get a straight line. Why?
Look at the equation: $V_0 = \left(\frac{h}{e}\right)\nu – \left(\frac{\Phi_0}{e}\right)$.
This perfectly matches the equation of a line $y = mx + c$.
– The slope of this graph is always $h/e$. (It is a universal constant, independent of the metal!)
– The X-intercept is the threshold frequency ($\nu_0$).
– The Y-intercept is $-\Phi_0/e$.
Look at the equation: $V_0 = \left(\frac{h}{e}\right)\nu – \left(\frac{\Phi_0}{e}\right)$.
This perfectly matches the equation of a line $y = mx + c$.
– The slope of this graph is always $h/e$. (It is a universal constant, independent of the metal!)
– The X-intercept is the threshold frequency ($\nu_0$).
– The Y-intercept is $-\Phi_0/e$.
Compare & Contrast
✗ Increasing Intensity
- Means increasing the number of photons hitting the metal per second.
- Increases the Photoelectric Current (more electrons get ejected).
- Does not change the Kinetic Energy of the electrons.
- Does not change the Stopping Potential.
✓ Increasing Frequency
- Means increasing the energy of each individual photon ($E = h\nu$).
- Increases the maximum Kinetic Energy of the emitted electrons.
- Increases (makes more negative) the Stopping Potential.
- Does not change the Photoelectric Current (saturation current remains the same).
Common Mistakes to Avoid
Mistake 1
Mismatched Units in Einstein’s Equation: You cannot subtract Work Function in $\text{eV}$ directly from $h\nu$ in Joules. You MUST convert one to match the other. To convert $\text{eV}$ to Joules, multiply by $1.6 \times 10^{-19}$. To convert Joules to $\text{eV}$, divide by $1.6 \times 10^{-19}$.
Mistake 2
Misunderstanding “Maximum” Kinetic Energy: The equation $K_{\text{max}} = h\nu – \Phi_0$ gives the energy of the fastest electrons (those emitted from the very surface). Electrons emitted from deeper inside the metal lose energy via collisions before escaping, so they come out with $K < K_{\text{max}}$. Not all electrons have the same energy!
Mistake 3
Applying $p = mv$ to Photons: Photons have zero rest mass, so you cannot use $p = mv$ or $K = \frac{1}{2}mv^2$ for them. For a photon, momentum is strictly $p = h/\lambda = E/c$. Save $p = mv$ for material particles like electrons or protons!
Exam Tips
Tip 1
The $1240$ Shortcut: When dealing with photons, calculating $E = hc/\lambda$ using standard SI units is tedious. Use the shortcut: $E (\text{in eV}) = \frac{1240 \text{ or } 1242}{\lambda (\text{in nm})}$. If wavelength is given in Angstroms (\AA), use $12400$. This saves massive amounts of calculation time in MCQ and numericals.
Tip 2
De Broglie Wavelength Ratios: When a particle is accelerated through potential $V$, $\lambda \propto 1/\sqrt{m q}$. If a proton and an alpha particle (mass $4m_p$, charge $2e$) are accelerated through the same potential, the ratio $\lambda_p / \lambda_\alpha = \sqrt{(m_\alpha q_\alpha) / (m_p q_p)} = \sqrt{4 \times 2} = \sqrt{8} = 2\sqrt{2}$.
Expected Exam Questions
SQ
Board Pattern Questions
Class 12 · Dual Nature · CBSE Exam1
The work function of a certain metal is $4.2 \text{ eV}$. Will this metal give photoelectric emission for incident radiation of wavelength $330 \text{ nm}$? [2 marks]
Answer
No, emission will not occur.
📝
Explanation
Use the shortcut to find the energy of the incident photon:
$E = \frac{1240}{\lambda \text{ (in nm)}} \text{ eV}$
$E = \frac{1240}{330} \approx 3.76 \text{ eV}$.
Since the incident energy ($3.76 \text{ eV}$) is less than the required work function ($4.2 \text{ eV}$), no photoelectric emission will take place.
2
An electron and a photon each have a wavelength of $1.00 \text{ nm}$. Find (i) their momenta, and (ii) the energy of the photon and the kinetic energy of the electron. [3 marks]
Answer
(i) $p = 6.63 \times 10^{-25} \text{ kg m/s}$, (ii) $E_\gamma = 1240 \text{ eV}$, $K_e \approx 1.5 \text{ eV}$
📝
Explanation
(i) Momenta: By De Broglie’s relation, if wavelengths are equal, momenta are equal.
$p = \frac{h}{\lambda} = \frac{6.63 \times 10^{-34}}{1 \times 10^{-9}} = 6.63 \times 10^{-25} \text{ kg m/s}$.
(ii) Energies:
For Photon: $E = \frac{hc}{\lambda} \text{ (shortcut)} = \frac{1240}{1 \text{ nm}} = 1240 \text{ eV}$.
For Electron: $K = \frac{p^2}{2m} = \frac{(6.63 \times 10^{-25})^2}{2 \times 9.1 \times 10^{-31}} = 2.41 \times 10^{-19} \text{ Joules}$.
To convert to eV: divide by $1.6 \times 10^{-19} \implies K \approx 1.5 \text{ eV}$.
Note: Even though they have the same wavelength and momentum, the photon has vastly more energy than the electron!
3
Draw a graph showing the variation of stopping potential with frequency of incident radiation for two different photosensitive materials having work functions $W_1$ and $W_2$ ($W_1 > W_2$). What does the slope of the graph represent? [3 marks]
Answer
Two parallel straight lines. Slope $= h/e$.
📝
Explanation
The graph consists of two parallel straight lines starting from different points on the positive x-axis (Threshold frequencies $\nu_1$ and $\nu_2$).
Since $W_1 > W_2$, the threshold frequency for material 1 is greater ($\nu_1 > \nu_2$), so its graph starts further to the right.
Because $V_0 = \left(\frac{h}{e}\right)\nu – \frac{W}{e}$, the slope of both lines is exactly $\frac{h}{e}$ (Planck’s constant divided by fundamental charge). Because $h$ and $e$ are universal constants, the lines must be parallel.
Concept Map
Dual Nature connects to →
Modern Physics
Atoms (Bohr Model depends on photons)
Wave Optics (Where wave theory failed)
Semiconductors (Photodiodes & Solar cells)
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