Photoelectric Effect & Matter Waves: Master Notes

Chapter Overview

The Photoelectric Effect established the particle nature of light, while de Broglie's hypothesis established the wave nature of matter. In IIT-JEE (Advanced), the most tested problems involve stopping potential ($V_0$) vs frequency ($\nu$) graphs, photoelectric current saturation vs intensity, matter wave wavelength under accelerating potentials ($\lambda =\frac{h}{\sqrt{2mqV}}$), and radiation pressure photon flux.

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1. Einstein's Photoelectric Equation

$$E_{\text{photon}}=h\nu =\frac{hc}{\lambda }\approx \frac{12400\text{eV}\cdot \text{AA}}{\lambda (\text{AA})}\approx \frac{1240\text{eV}\cdot \text{nm}}{\lambda (\text{nm})}$$$$\boxed{{\displaystyle h\nu =\mathrm{\Phi }+K_{max}=\mathrm{\Phi }+e{V}_0}}$$

Where:

• $\mathrm{\Phi }=h{\nu }_0=\frac{hc}{{\lambda }_0}$ is the Work Function of the photosensitive metal.
• ${\nu }_0$ is the Threshold Frequency, ${\lambda }_0$ is the Threshold Wavelength.
• $V_0$ is the Stopping Potential (retarding voltage that drops photocurrent to zero).

Multi-Mode DiagramRotational Dynamics: Spool on Incline & Pure Rolling

Option 1: Publication-Grade Scientific Vector SVG

Free-body force decomposition on incline showing normal force N, static friction fs, tension T, and Instantaneous Axis of Rotation C.

Linear Acceleration: $a_{\text{cm}} = \frac{g\sin\theta}{1 + I_{\text{cm}}/(MR^2)} = \frac{2}{3}g\sin\theta$

Critical Tension Angle: $\phi_c = \arccos(r/R)$

Option 2: Interactive 60 FPS Canvas Physics Engine

Real-time 60 FPS pure rolling simulation on incline.

Incline Angle θ: 25 °

Acceleration ($a_{\text{cm}}$):2.76 m/s²

Option 3: Interactive 3D Spatial Viewport

3D rotatable gyroscope disc.

+X

+Y

+Z

👆 Click and drag anywhere to rotate in 3D space

Azimuth Angle (ψ):35°

Elevation Angle (θ):-22°

Precession Equation:$\vec{\tau} = \frac{d\vec{L}}{dt}$

Option 5: Reactive Parametric Physics with Live Calculus

Parametric angle sliders with live torque calculation.

Incline Angle θ: 25 °

Pull Angle φ: 45 °

Critical Angle ($\phi_c$):$62.5°$

Active Mode:Option 1 (Print & Derivation Vector SVG)Switch modes using the tabs above to explore different visualizations.

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2. Photocurrent Characteristics & Photon Flux

• Photon Flux ($N_{\text{ph}}$): Number of photons arriving per unit time on area $A$:$$N_{\text{ph}}=\frac{P_{\text{power}}}{h\nu }=\frac{I_{\text{intensity}}A}{h\nu }$$
• Saturation Photocurrent ($I_{\text{sat}}$):$$I_{\text{sat}}=\eta \cdot e\cdot N_{\text{ph}}=\eta e\left(\frac{P}{h\nu }\right)$$(where $\eta$ is the quantum efficiency, typically $1\mathrm{\%}-5\mathrm{\%}$).
• Key Experimental Laws:
  1. Saturation current is strictly proportional to light intensity ($I_{\text{sat}}\propto \text{Intensity}$).
  2. Stopping potential $V_0$ is independent of light intensity and depends linearly on frequency $\nu$.
  3. Photoelectric emission is instantaneous (time lag $<{10}^{-9}\text{ s}$).

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3. de Broglie Matter Waves

$$\boxed{{\displaystyle \lambda =\frac{h}{p}=\frac{h}{mv}=\frac{h}{\sqrt{2mK}}}}$$

3.1 Standard de Broglie Wavelengths under Acceleration $V$

Particle	Charge & Mass	Wavelength Formula $\lambda$
Electron	$q=e,m=m_e$	${\lambda }_e=\frac{h}{\sqrt{2m_{e}eV}}=\boxed{{\displaystyle \sqrt{\frac{150}{V}}\text{AA}\approx \frac{12.27}{\sqrt{V}}\text{AA}}}$
Proton	$q=e,m=m_p$	${\lambda }_p=\frac{0.286}{\sqrt{V}}\text{AA}$
Alpha Particle (${\text{He}}^{2+}$)	$q=2e,m=4m_p$	${\lambda }_{\alpha }=\frac{0.101}{\sqrt{V}}\text{AA}$
Thermal Gas Particle (Temp $T$)	$K=\frac{3}{2}{k}_{B}T$	${\lambda }_{\text{thermal}}=\frac{h}{\sqrt{3m{k}_{B}T}}$

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4. Authentic Previous Years Questions (PYQs)

PYQ 1: JEE Advanced 2022 (Paper 1) — Photoelectric Stopping Potential Graph

Question:
When light of frequency ${\nu }_1=8.0\times {10}^{14}\text{ Hz}$ illuminates a metal surface, the stopping potential is $V_{01}=2.0\text{ V}$. When light of frequency ${\nu }_2=4.0\times {10}^{14}\text{ Hz}$ illuminates the same metal, the stopping potential is $V_{02}=0.5\text{ V}$. Find:

1. Planck's constant $h$.
2. The work function $\mathrm{\Phi }$ of the metal in $\text{eV}$.

Step-by-Step Solution:

• 1. Using Einstein's Equation for Both Frequencies:

  $$e{V}_{01}=h{\nu }_1-\mathrm{\Phi }\text{--- (1)}$$$$e{V}_{02}=h{\nu }_2-\mathrm{\Phi }\text{--- (2)}$$

  Subtracting (2) from (1):

  $$e(V_{01}-V_{02})=h({\nu }_1-{\nu }_2)$$$$h=\frac{e(2.0-0.5)}{8.0\times {10}^{14}-4.0\times {10}^{14}}=\frac{1.6\times {10}^{-19}\times 1.5}{4.0\times {10}^{14}}=\frac{2.4\times {10}^{-19}}{4.0\times {10}^{14}}=\boxed{{\displaystyle 6.0\times {10}^{-34}\text{ J}\cdot \text{s}}}$$

• 2. Work Function $\mathrm{\Phi }$:

  $$\mathrm{\Phi }=h{\nu }_2-e{V}_{02}=\frac{h{\nu }_2}{e}-V_{02}=\frac{(6.0\times {10}^{-34})(4.0\times {10}^{14})}{1.6\times {10}^{-19}}\text{ eV}-0.5\text{ eV}$$$$\mathrm{\Phi }=\frac{2.4\times {10}^{-19}}{1.6\times {10}^{-19}}\text{ eV}-0.5\text{ eV}=1.5\text{ eV}-0.5\text{ eV}=\boxed{{\displaystyle 1.0\text{ eV}}}$$

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PYQ 2: JEE Advanced 2020 — de Broglie Wavelength Ratio

Question:
An electron and a photon each have a de Broglie wavelength of $\lambda =1.0\text{ nm}$. Find the ratio of the kinetic energy of the photon to the kinetic energy of the electron. ($m_e=9.1\times {10}^{-31}\text{ kg}$).

Step-by-Step Solution:

1. Energy of Photon:$$E_{\text{photon}}=\frac{hc}{\lambda }$$
2. Kinetic Energy of Electron:$$K_{\text{electron}}=\frac{p^2}{2m_e}=\frac{h^2}{2m_e{\lambda }^2}$$
3. Ratio:$$\frac{E_{\text{photon}}}{K_{\text{electron}}}=\frac{\frac{hc}{\lambda }}{\frac{h^2}{2m_e{\lambda }^2}}=\frac{2m_{e}c\lambda }{h}$$$$\frac{E_{\text{photon}}}{K_{\text{electron}}}=\frac{2\times (9.1\times {10}^{-31})\times (3\times {10}^8)\times (1.0\times {10}^{-9})}{6.63\times {10}^{-34}}=\frac{5.46\times {10}^{-31}}{6.63\times {10}^{-34}}\approx \boxed{{\displaystyle 823.5}}$$

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5. High-Yield Formula Sheet

Entity	Formula	Notes
Einstein Photoelectric	$e{V}_0=h\nu -\mathrm{\Phi }$	Linear $V_0$ vs $\nu$
Electron Wavelength	${\lambda }_e=\sqrt{\frac{150}{V}}\text{AA}$	Accelerated through $V$ volts
Thermal Gas de Broglie	$\lambda =\frac{h}{\sqrt{3m{k}_{B}T}}$	Root mean square momentum
Photon Energy Shortcut	$E(\text{eV})=\frac{1240}{\lambda (\text{nm})}$	Rapid exam calculation