🔙 Prerequisite Revision:
👉 Chapter 10: Wave Optics (Interference & Diffraction)Chapter 11: Dual Nature of Radiation and Matter
Vol 1: Electron Emission & Photoelectric Effect
Free Electrons: In metals, electrons in the outer shell are loosely bound. However, they cannot leave the metal surface on their own due to the attractive pull of the nucleus.
• Unit: Electron Volt (eV). ($1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$).
• Cesium (Cs) has the lowest work function ($2.14$ eV).
• Platinum (Pt) has the highest work function ($5.65$ eV).
The Photoelectric Effect
The phenomenon of emission of electrons from a metal surface when light (radiation) of suitable frequency falls on it.
- Hertz's Observation: Spark discharge occurred more easily when illuminated by UV light.
- Hallwachs & Lenard: Confirmed that ejected particles are negatively charged electrons.
- Threshold Frequency ($\nu_0$): The minimum frequency of incident light required to cause photo-emission. If $\nu < \nu_0$, no emission occurs, no matter how high the intensity.
Vol 2: Einstein's Photoelectric Equation
Albert Einstein (1905) explained the photoelectric effect using Planck's Quantum Theory (Particle nature of light).
Light consists of discrete packets of energy called Photons ($E = h\nu$).
Energy of Photon = Work Function + Max Kinetic Energy
$$h\nu = \Phi_0 + K_{max}$$
Or: $$K_{max} = h\nu - h\nu_0$$
Stopping Potential ($V_0$)
The minimum negative (retarding) potential applied to the anode that stops even the most energetic photoelectrons.
Important Graphs (Exam Favorites)
From Einstein's equation: $eV_0 = h\nu - \Phi_0 \Rightarrow V_0 = (\frac{h}{e})\nu - \frac{\Phi_0}{e}$.
This is a straight line equation ($y = mx + c$).
• Slope: $h/e$ (Universal Constant).
• Intercept: $-\Phi_0/e$.
Vol 3: Wave Nature of Matter (De-Broglie)
De-Broglie Hypothesis (1924): Since nature loves symmetry, if radiation has a dual nature, then matter (particles like electrons) should also possess a wave nature.
$$\lambda = \frac{h}{p} = \frac{h}{mv}$$
Wavelength of an Electron
If an electron is accelerated through a potential difference of $V$ volts, its Kinetic Energy $K = eV$.
Momentum $p = \sqrt{2mK} = \sqrt{2meV}$.
Substituting values ($h, m, e$):
$$\lambda = \frac{1.227}{\sqrt{V}} \text{ nm}$$
Davisson-Germer Experiment: It experimentally verified the wave nature of electrons by observing diffraction of electrons by nickel crystals.
Vol 4: Mission 100 Question Bank
Section A: MCQs
Q1. The slope of the graph between stopping potential ($V_0$) and frequency ($\nu$) is:
(a) $h$ (b) $h/e$ (c) $e/h$ (d) $\Phi_0$
Ans: (b) $h/e$.
Q2. Which photon has more energy?
(a) Red (b) Violet (c) Yellow (d) Green
Ans: (b) Violet ($E = h\nu$, frequency of violet is highest).
Section B: Numericals (Important)
Q3. Calculate the de-Broglie wavelength of an electron accelerated through a potential of 100 V.
Solution:
Formula: $\lambda = \frac{1.227}{\sqrt{V}}$ nm
$\lambda = \frac{1.227}{\sqrt{100}} = \frac{1.227}{10}$ nm
$\lambda = 0.1227$ nm (or $1.227 \text{ \AA}$).
Ans: 0.123 nm.
Q4. The work function of Cesium is 2.14 eV. Find the threshold frequency.
Solution:
Given: $\Phi_0 = 2.14 \text{ eV} = 2.14 \times 1.6 \times 10^{-19}$ J.
Formula: $\Phi_0 = h\nu_0 \Rightarrow \nu_0 = \Phi_0/h$
$\nu_0 = \frac{3.424 \times 10^{-19}}{6.63 \times 10^{-34}}$
$\nu_0 \approx 5.16 \times 10^{14}$ Hz.
Ans: $5.16 \times 10^{14}$ Hz.
Section C: Conceptual
Q5. Why did the Wave Theory of light fail to explain the Photoelectric Effect?
Ans: According to Wave Theory, energy is distributed continuously. It predicted that electron emission would take time (time lag) and any frequency could cause emission if intensity is high enough. Both were proved wrong experimentally.
Mission 100 Physics Series
Next Chapter: Atoms & Nuclei (The Core of Matter!)
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