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AXIOM PHYSICS Class 12 • Open Access

CBSE Class XII • Unit 07 (Modern Physics)

Chapter 11: Dual Nature of Radiation and Matter

Master Revision Dossier • Photoelectric Effect, Einstein's Equation, Millikan's Slope ($h/e$), Photon Properties & de Broglie Matter Waves

1. Definitive Formula & Quantum Quantities Matrix

Concept Mathematical Formulation SI Unit / Value Boundary Conditions & Notes
Work Function $$\Phi_0 = h\nu_0 = \frac{hc}{\lambda_0}$$ $\text{eV}$ or $\text{J}$
($1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$)		 Minimum energy required to liberate electron from metal surface. Lowest for Cesium ($2.14\text{ eV}$); highest for Platinum ($5.65\text{ eV}$).
Einstein's Equation $$K_{\max} = h\nu - \Phi_0 = h(\nu - \nu_0)$$ $K_{\max} = \frac{1}{2}m v_{\max}^2$ Derived via 1-to-1 photon-electron interaction. Independent of light intensity. If $\nu < \nu_0$, no emission occurs regardless of intensity.
Stopping Potential $$e V_0 = K_{\max} \implies V_0 = \left(\frac{h}{e}\right)\nu - \frac{\Phi_0}{e}$$ $\text{Volts (V)}$ Plot of $V_0 \text{ vs } \nu$ is a straight line with universal slope $\frac{h}{e} \approx 4.14 \times 10^{-15}\text{ V}\cdot\text{s}$ (identical for all metals).
Photon Momentum $$p = \frac{E}{c} = \frac{h\nu}{c} = \frac{h}{\lambda}$$ $\text{kg}\cdot\text{m/s}$ Photon rest mass is zero ($m_0 = 0$). Radiation force: $F = \frac{IA}{c}$ (absorbed), $F = \frac{2IA}{c}$ (reflected).
de Broglie Relation $$\lambda = \frac{h}{p} = \frac{h}{mv} = \frac{h}{\sqrt{2mK}}$$ $\text{Metres (m)}$ Universal wave-particle duality. For charged particles accelerated through potential $V$: $K = qV \implies \lambda = \frac{h}{\sqrt{2mqV}}$.
Electron Wavelength $$\lambda_e = \frac{1.227}{\sqrt{V}}\text{ nm} = \frac{12.27}{\sqrt{V}}\text{ \AA}$$ $V \text{ in Volts}$ For $V = 100\text{ V} \implies \lambda_e = 1.227\text{ \AA}$ (matches X-ray regime and interatomic spacing in crystals, allowing electron diffraction).
Thermal Neutrons $$\lambda_n = \frac{h}{\sqrt{3mkT}}$$ $T \text{ in Kelvin (K)}$ Average kinetic energy of thermal neutron: $K = \frac{3}{2}kT$. At $T = 300\text{ K}$, $\lambda_n \approx 0.145\text{ nm} = 1.45\text{ \AA}$.

2. Core Derivations Blueprint (Step-by-Step Board Proofs)

Blueprint 11.1: Deduction of Einstein's Equation & Determination of Planck's Constant ($h$)

1. According to Einstein's quantum postulate, incident light delivers energy in discrete packets (photons) of energy $E = h\nu$.

2. When a photon strikes an electron on a metal surface, it transfers its entire energy instantaneously. Work function $\Phi_0$ is consumed to liberate the electron, and the remainder appears as maximum kinetic energy $K_{\max}$:

$$h\nu = \Phi_0 + K_{\max} \implies K_{\max} = h\nu - \Phi_0 = h(\nu - \nu_0)$$

3. If $V_0$ is the stopping potential: $K_{\max} = e V_0 \implies e V_0 = h\nu - \Phi_0$.

$$V_0 = \left(\frac{h}{e}\right)\nu - \frac{\Phi_0}{e}$$

4. Comparing with $y = mx + c$, the graph of $V_0$ versus $\nu$ is a straight line:

$$\text{Slope } m = \frac{\Delta V_0}{\Delta \nu} = \frac{h}{e} \implies h = e \times (\text{Slope})$$

5. Intercept on frequency axis ($\nu$-axis): $V_0 = 0 \implies \nu = \nu_0 = \frac{\Phi_0}{h}$ (Threshold Frequency). Intercept on potential axis: $c = -\frac{\Phi_0}{e}$.

Blueprint 11.2: Derivation of de Broglie Wavelength of an Accelerated Electron

1. An electron of mass $m$ and charge $e$ is accelerated from rest through a potential difference of $V$ volts.

2. Work done on electron equals kinetic energy: $K = e V$.

3. Momentum of electron: $K = \frac{p^2}{2m} \implies p = \sqrt{2mK} = \sqrt{2m e V}$.

4. de Broglie wavelength: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m e V}}$.

5. Substituting $h = 6.626 \times 10^{-34}\text{ J}\cdot\text{s}$, $m = 9.109 \times 10^{-31}\text{ kg}$, $e = 1.602 \times 10^{-19}\text{ C}$:

$$\lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times (9.109 \times 10^{-31}) \times (1.602 \times 10^{-19}) \times V}} = \frac{6.626 \times 10^{-34}}{5.402 \times 10^{-25}\sqrt{V}}$$
$$\lambda = \frac{1.227 \times 10^{-9}}{\sqrt{V}}\text{ m} = \frac{1.227}{\sqrt{V}}\text{ nm} = \frac{12.27}{\sqrt{V}}\text{ \AA}$$

3. High-Yield "Red Alert" Exam Trap Matrix

Master Trap 1: de Broglie Wavelength Ratios for Charged Particles

Particle Pair Same Potential ($V$) Formula Same Kinetic Energy ($K$) Formula Numerical Ratio
Proton ($p$) & Alpha ($\alpha$) $$\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha q_\alpha}{m_p q_p}} = \sqrt{\frac{4 \times 2}{1 \times 1}}$$ $$\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha}{m_p}} = \sqrt{\frac{4}{1}}$$ Same $V \implies \mathbf{2\sqrt{2} \approx 2.83}$
Same $K \implies \mathbf{2:1}$
Deuteron ($d$) & Alpha ($\alpha$) $$\frac{\lambda_d}{\lambda_\alpha} = \sqrt{\frac{m_\alpha q_\alpha}{m_d q_d}} = \sqrt{\frac{4 \times 2}{2 \times 1}}$$ $$\frac{\lambda_d}{\lambda_\alpha} = \sqrt{\frac{m_\alpha}{m_d}} = \sqrt{\frac{4}{2}}$$ Same $V \implies \mathbf{2:1}$
Same $K \implies \mathbf{\sqrt{2} \approx 1.414}$
Proton ($p$) & Deuteron ($d$) $$\frac{\lambda_p}{\lambda_d} = \sqrt{\frac{m_d q_d}{m_p q_p}} = \sqrt{\frac{2 \times 1}{1 \times 1}}$$ $$\frac{\lambda_p}{\lambda_d} = \sqrt{\frac{m_d}{m_p}} = \sqrt{\frac{2}{1}}$$ Same $V \implies \mathbf{\sqrt{2} \approx 1.414}$
Same $K \implies \mathbf{\sqrt{2} \approx 1.414}$

⚠️ Trap 2: Slope of $V_0 \text{ vs } \nu$ is Universal

The slope of stopping potential versus frequency is strictly $h/e$ for ALL metals.
• Changing the metal only shifts the intercept ($\nu_0$ and $-\Phi_0/e$).
• The lines for different metals are strictly PARALLEL.

⚠️ Trap 3: Doubling Frequency Effect on $K_{\max}$

If incident frequency $\nu$ is doubled:
$$K_{\max}' = h(2\nu) - \Phi_0 = 2(h\nu - \Phi_0) + \Phi_0 = \mathbf{2K_{\max} + \Phi_0}$$ Therefore, doubling the frequency makes $K_{\max}$ MORE THAN DOUBLE!

4. Handcrafted Exemplars (Analytical Numerical Solutions)

Problem 1: Stopping Potential & Work Function Determination Exemplar 01

When light of frequency $8.2 \times 10^{14}\text{ Hz}$ falls on a photosensitive surface, the stopping potential is $1.25\text{ V}$. When light of frequency $12.0 \times 10^{14}\text{ Hz}$ falls on the same surface, the stopping potential is $2.82\text{ V}$. Calculate Planck's constant and the work function of the metal.

Solution: From $e(V_{02} - V_{01}) = h(\nu_2 - \nu_1)$:

$$h = \frac{e(V_{02} - V_{01})}{\nu_2 - \nu_1} = \frac{(1.6 \times 10^{-19}) \times (2.82 - 1.25)}{(12.0 - 8.2) \times 10^{14}} = \frac{1.6 \times 10^{-19} \times 1.57}{3.8 \times 10^{14}} = \frac{2.512 \times 10^{-19}}{3.8 \times 10^{14}}$$
$$h \approx 6.61 \times 10^{-34}\text{ J}\cdot\text{s}$$

Work function:

$$\Phi_0 = h\nu_1 - e V_{01} = \frac{(6.61 \times 10^{-34}) \times (8.2 \times 10^{14})}{1.6 \times 10^{-19}}\text{ eV} - 1.25\text{ eV} = 3.39\text{ eV} - 1.25\text{ eV} = 2.14\text{ eV} \quad (\text{Cesium})$$
Problem 2: Ratio of de Broglie Wavelengths Exemplar 02

An electron, an alpha particle, and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength? What if they are accelerated through the same potential difference $V$?

Case (a) Same Kinetic Energy: $\lambda = \frac{h}{\sqrt{2mK}} \implies \lambda \propto \frac{1}{\sqrt{m}}$.

Since $m_\alpha > m_p > m_e$, the particle with largest mass has the shortest wavelength: $\alpha$-particle has the shortest wavelength.

Case (b) Same Potential $V$: $\lambda = \frac{h}{\sqrt{2mqV}} \implies \lambda \propto \frac{1}{\sqrt{mq}}$.

Product $m q$: For electron $= m_e e$; for proton $= m_p e$; for alpha $= (4m_p)(2e) = 8 m_p e$.

$$\text{Since } (mq)_\alpha \text{ is maximum, the } \mathbf{\alpha}\text{\textbf{-particle again has the shortest wavelength!}}$$

5. Printable Pocket Flashcard Grid

Einstein's Law

$K_{\max} = h\nu - \Phi_0$

Stopping Slope

$\frac{\Delta V_0}{\Delta\nu} = \frac{h}{e}$

Electron Wavelength

$\lambda_e = \frac{1.227}{\sqrt{V}}\text{ nm}$

Proton / Alpha Ratio

$\frac{\lambda_p}{\lambda_\alpha} = 2\sqrt{2}$