Axiom Physics | Class 12 - Einstein's Photoelectric Equation & Photon Picture

⚡ 01. The Spark (Analogy)

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Albert Einstein ne 1905 mein yahi simple thermodynamic logic use kiya: Light continuous stream nahi hai, balki energy ke chhote-chhote packets (quanta) yaani Photons ki bauchhar hai. Jab ek photon kisi surface electron se takrata hai, toh woh apni poori energy ($E = h\nu$) us ek electron ko dekar khud gayab ho jata hai (1-to-1 collision). Us energy ka ek hissa electron ko metal ki jail se azaad karane mein kharch hota hai (Work Function $\Phi_0$), aur bachi hui saari energy electron ki daudne ki raftar ban jaati hai (Maximum Kinetic Energy $K_{\max}$)!

📐 02. The Core Law & Formulae

1. Einstein's Photoelectric Equation

Conservation of energy for single photon-electron interaction:

$$K_{\max} = h\nu - \Phi_0 = h(\nu - \nu_0) = hc\left(\frac{1}{\lambda} - \frac{1}{\lambda_0}\right)$$

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

Equation of a straight line ($y = mx + c$) with slope $m = h/e$ (Universal Constant) and intercept $c = -\Phi_0/e$ on the potential axis.

2. The 5 Cardinal Photon Characteristics

Energy: $E = h\nu = \frac{hc}{\lambda}$ (Depends purely on frequency).
Momentum: $p = \frac{E}{c} = \frac{h\nu}{c} = \frac{h}{\lambda}$.
Rest Mass: Rest mass of a photon is strictly ZERO ($m_0 = 0$).
Electrical Neutrality: Photons are unaffected by electric and magnetic fields.
Collision Invariant: In photon-particle collisions, total energy and momentum are conserved, but number of photons may not be conserved (photons can be absorbed or created).

03. Interactive Visualizer: Millikan's Linear $V_0 \text{ vs } \nu$ Slope Engine

Incident Frequency ($\nu$) 8.50 × 10¹⁴ Hz

Target Metal Work Function ($\Phi_0$) 2.14 eV (Cesium)

Preset Target Elements Preset Selector

Slope of $V_0-\nu$ graph ($h/e$):

4.136 × 10⁻¹⁵ V·s

Threshold Frequency ($\nu_0$):

5.17 × 10¹⁴ Hz

Stopping Potential ($V_0$):

+1.38 V

Planck's Constant ($h$):

6.626 × 10⁻³⁴ J·s

📝 04. Comprehensive Stepwise Derivations Blueprint

Primary Board Derivation

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

Fig 11.2A: Millikan's experimental verification of Einstein's photoelectric equation; all metals exhibit identical universal slope $h/e$.

1. According to Einstein's quantum postulate, radiation of frequency $\nu$ consists of discrete quanta (photons), each possessing energy $E = h\nu$.

2. When a photon collides with an electron on the metal surface, it transfers its entire energy $h\nu$ instantaneously. A minimum energy $\Phi_0$ (Work Function) is consumed to overcome the surface nuclear barrier. The remaining energy appears as the maximum kinetic energy $K_{\max}$ of the emitted photoelectron:

$$E = \Phi_0 + K_{\max} \implies K_{\max} = h\nu - \Phi_0$$

3. If the incident frequency equals the threshold frequency $\nu_0$, electrons are emitted with zero kinetic energy ($K_{\max} = 0 \implies \Phi_0 = h\nu_0$):

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

4. Since the stopping potential $V_0$ stops the fastest electrons ($K_{\max} = e V_0$):

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

5. Comparing with standard straight-line equation $y = mx + c$:

• Slope of the line ($m$): $\text{Slope} = \frac{h}{e}$. Since both $h$ and $e$ are fundamental physical constants, the slope is a universal constant identical for all metals!

• Determination of Planck's Constant: By experimentally measuring the slope of the $V_0 \text{ vs } \nu$ plot and multiplying it by electronic charge $e$ ($1.6 \times 10^{-19}\text{ C}$), Robert Millikan precisely determined Planck's constant:

$$h = e \times \text{Slope} = e \times \left(\frac{\Delta V_0}{\Delta \nu}\right) \approx 6.626 \times 10^{-34}\text{ J}\cdot\text{s}$$

• Intercept on Frequency Axis ($V_0 = 0$): $\nu = \nu_0 = \frac{\Phi_0}{h}$ (Threshold frequency).

• Intercept on Potential Axis ($\nu = 0$): $c = -\frac{\Phi_0}{e}$ (Direct measurement of work function in Volts).

Quantum Momentum

Photon Momentum, Force & Radiation Pressure Formulations

1. By Einstein's mass-energy equivalence ($E = mc^2$) and de Broglie relation:

$$E = pc \implies p = \frac{E}{c} = \frac{h\nu}{c} = \frac{h}{\lambda}$$

2. Let a light beam of power $P$ and intensity $I = P/A$ be incident on a surface of area $A$. Number of photons incident per second ($n$):

$$n = \frac{P}{E} = \frac{I A}{h\nu}$$

3. Case (a) Completely Absorbing Surface: Every photon delivers initial momentum $p = \frac{h\nu}{c}$ and is absorbed ($p_f = 0 \implies \Delta p = p$):

$$\text{Force } F = n \cdot \Delta p = \left(\frac{I A}{h\nu}\right)\left(\frac{h\nu}{c}\right) = \frac{I A}{c}, \qquad \text{Pressure } P_{\text{rad}} = \frac{F}{A} = \frac{I}{c}$$

4. Case (b) Completely Reflecting Surface (Perfect Mirror): Photons bounce back with inverted momentum ($p_f = -p \implies \Delta p = 2p$):

$$\text{Force } F = \frac{2 I A}{c}, \qquad \text{Pressure } P_{\text{rad}} = \frac{2I}{c}$$

⚠️ 05. Red Alert Exam Trap Zone

Trap 1: The "Slope Depends on Target Metal" Myth

*Board Trap Question:* If we plot $V_0 \text{ vs } \nu$ for Sodium ($\Phi_0 = 2.46\text{ eV}$) and Platinum ($\Phi_0 = 5.65\text{ eV}$), which line will have a steeper slope?
*Correct Answer:* BOTH HAVE THE EXACT SAME SLOPE! Slope is strictly equal to $h/e$, which depends neither on the nature of the metal nor on the intensity of light. The lines are strictly PARALLEL to each other!

Trap 2: $K_{\max} \text{ vs } \nu$ Slope vs $V_0 \text{ vs } \nu$ Slope

Read the vertical axis label with eagle eyes:
• If y-axis is Stopping Potential ($V_0$): $\text{Slope} = \mathbf{h/e}$.
• If y-axis is Maximum Kinetic Energy ($K_{\max}$): $\text{Slope} = \mathbf{h}$ (Planck's constant directly!).

Trap 3: Doubling Intensity vs Doubling Frequency

• If intensity is doubled (at fixed $\nu$): Photoelectric current doubles; $K_{\max}$ and $V_0$ remain completely unchanged.
• If frequency $\nu$ is doubled: $K_{\max} = h(2\nu) - \Phi_0 = 2(h\nu - \Phi_0) + \Phi_0 = 2K_{\max} + \Phi_0$. Therefore, doubling the frequency makes $K_{\max}$ MORE THAN DOUBLE!

Trap 4: Are All Emitted Electrons Possessing $K_{\max}$?

*Conceptual Trap:* Do all photoelectrons leave the metal surface with identical kinetic energy $K_{\max}$?
*Correct Answer:* NO! $K_{\max}$ is possessed only by electrons knocked off from the outermost surface layer. Electrons knocked out from deeper atomic layers lose energy via internal collisions before reaching the surface. The kinetic energies of emitted electrons form a continuous range: $$\mathbf{0 \le K \le K_{\max}}$$

💡 06. Handcrafted Exemplars

Board Classic Level

Problem: The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength $491\text{ nm}$ is $0.710\text{ V}$. When the incident wavelength is changed to $388\text{ nm}$, the new stopping potential is found to be $1.43\text{ V}$. Calculate the experimental value of Planck’s constant and the work function of the target metal.

Solution: From Einstein's equation: $e V_0 = \frac{hc}{\lambda} - \Phi_0$. Writing for both wavelengths:

$$e V_{01} = \frac{hc}{\lambda_1} - \Phi_0 \quad \text{--- [Eq. 1]}$$ $$e V_{02} = \frac{hc}{\lambda_2} - \Phi_0 \quad \text{--- [Eq. 2]}$$

Subtracting Eq. 1 from Eq. 2:

$$e (V_{02} - V_{01}) = hc \left(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}\right) = hc \left(\frac{\lambda_1 - \lambda_2}{\lambda_1 \lambda_2}\right)$$

Substitute values: $\Delta V_0 = 1.43 - 0.710 = 0.720\text{ V}$, $\lambda_1 = 491 \times 10^{-9}\text{ m}, \lambda_2 = 388 \times 10^{-9}\text{ m}$:

$$(1.6 \times 10^{-19}) \times 0.720 = h \times (3 \times 10^8) \times \left[ \frac{(491 - 388) \times 10^{-9}}{(491 \times 10^{-9}) \times (388 \times 10^{-9})} \right]$$ $$1.152 \times 10^{-19} = h \times (3 \times 10^8) \times \left[ \frac{103}{1.905 \times 10^{-13}} \right] = h \times 1.622 \times 10^{23}$$

$$h = \frac{1.152 \times 10^{-19}}{1.73 \times 10^{14}} \approx 6.65 \times 10^{-34}\text{ J}\cdot\text{s}$$

Finding Work Function from Eq. 1:

$$\Phi_0 = \frac{hc}{\lambda_1} - e V_{01} = \frac{(6.65 \times 10^{-34}) \times (3 \times 10^8)}{491 \times 10^{-9} \times 1.6 \times 10^{-19}}\text{ eV} - 0.710\text{ eV} = 2.54\text{ eV} - 0.71\text{ eV} = 1.83\text{ eV}$$

JEE/NEET Competitive Trap

Problem: A monochromatic point source of light of power $20\text{ W}$ emits light of frequency $6 \times 10^{14}\text{ Hz}$. Find (a) the energy of each photon in eV, (b) the momentum of each photon, and (c) the number of photons emitted per second by the source.

Solution: Power $P = 20\text{ W} = 20\text{ J/s}$, $\nu = 6.0 \times 10^{14}\text{ Hz}$.

1. Energy of each photon:

$$E = h\nu = (6.63 \times 10^{-34}\text{ J}\cdot\text{s}) \times (6.0 \times 10^{14}\text{ s}^{-1}) = 3.978 \times 10^{-19}\text{ Joules}$$

$$E = \frac{3.978 \times 10^{-19}\text{ J}}{1.6 \times 10^{-19}\text{ J/eV}} \approx 2.49\text{ eV}$$

2. Momentum of each photon ($p = E/c$):

$$p = \frac{3.978 \times 10^{-19}\text{ J}}{3.0 \times 10^8\text{ m/s}} = 1.326 \times 10^{-27}\text{ kg}\cdot\text{m/s}$$

3. Number of photons emitted per second ($N = P/E$):

$$N = \frac{20\text{ J/s}}{3.978 \times 10^{-19}\text{ J}} \approx 5.03 \times 10^{19}\text{ photons/second}$$

🗂️ 07. Quick Revision Flashcard Core Equations

Einstein's Equation

$K_{\max} = h\nu - \Phi_0 = eV_0$

Universal Slope

$\frac{\Delta V_0}{\Delta\nu} = \frac{h}{e}$ (Same for all)

Photon Momentum

$p = \frac{h\nu}{c} = \frac{h}{\lambda}$

Photon Rest Mass

$m_0 = 0$ (Invariant)

Einstein's Equation, Planck's Constant & Photon Picture