Photoelectric Effect

Introduction

The photoelectric effect is the emission of electrons from a metal surface when light of sufficient frequency shines on it. This phenomenon provided crucial evidence for the particle nature of light and earned Einstein the Nobel Prize in 1921.

Historical Context

Classical wave theory predicted:

• Higher intensity should give electrons more energy
• Any frequency should eventually cause emission

Experiments showed otherwise! This led to the quantum explanation.

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Experimental Observations

Key Findings

1. Threshold frequency ($\nu_0$): Below this frequency, no electrons are emitted regardless of intensity

2. Instantaneous emission: Electrons are emitted immediately (within 10⁻⁹ s), not gradually

3. Intensity effect: More intensity = more electrons, but NOT higher kinetic energy

4. Frequency effect: Higher frequency = higher kinetic energy of emitted electrons

graph TD
    A[Light hits metal] --> B{ν ≥ ν₀?}
    B -->|Yes| C[Electrons emitted]
    B -->|No| D[No emission]
    C --> E[KE depends on frequency]
    C --> F[Number depends on intensity]

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Einstein’s Explanation

Einstein (1905) proposed that light consists of photons, each carrying energy $E = h\nu$.

The Photoelectric Equation

$$\boxed{h\nu = \phi + \frac{1}{2}mv_{max}^2}$$

or equivalently:

$$\boxed{h\nu = h\nu_0 + KE_{max}}$$

where:

• $h\nu$ = Energy of incident photon
• $\phi = h\nu_0$ = Work function (minimum energy to remove electron)
• $\frac{1}{2}mv_{max}^2 = KE_{max}$ = Maximum kinetic energy of electron
• $\nu_0$ = Threshold frequency

Interactive Demo: Visualize Photoelectric Effect

See how light photons eject electrons from a metal surface based on frequency and intensity.

Understanding Work Function

The work function ($\phi$) is the minimum energy needed to remove an electron from the metal surface.

Metal	Work Function (eV)
Cesium (Cs)	2.14
Potassium (K)	2.30
Sodium (Na)	2.75
Zinc (Zn)	4.33
Platinum (Pt)	5.65

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Stopping Potential

When a negative potential is applied to the collector, it can stop the most energetic electrons.

Definition

Stopping potential ($V_0$): The minimum negative potential needed to stop all photoelectrons.

$$\boxed{eV_0 = KE_{max} = h\nu - \phi}$$

or:

$$V_0 = \frac{h\nu - \phi}{e} = \frac{h(\nu - \nu_0)}{e}$$

Graphical Analysis

Plotting $V_0$ vs $\nu$:

• Slope = $\frac{h}{e}$ (universal constant)
• x-intercept = threshold frequency $\nu_0$
• y-intercept = $-\frac{\phi}{e}$

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Threshold Wavelength

Since $\nu_0 = \frac{c}{\lambda_0}$, we can also define:

$$\boxed{\lambda_0 = \frac{hc}{\phi}}$$

• If $\lambda < \lambda_0$: emission occurs
• If $\lambda > \lambda_0$: no emission

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Problem Solving Strategy

Step-by-step Approach

1. Identify given values: frequency/wavelength, work function, KE
2. Convert units: Ensure consistent units (eV or J)
3. Apply Einstein’s equation: $h\nu = \phi + KE_{max}$
4. Solve for unknown

Example 1: Finding KE

Problem: Light of wavelength 200 nm falls on a metal with work function 4.2 eV. Find the maximum KE of photoelectrons.

Solution:

1. Energy of photon:

   $$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{200 \times 10^{-9}} = 9.94 \times 10^{-19} \text{ J}$$ $$E = \frac{9.94 \times 10^{-19}}{1.6 \times 10^{-19}} = 6.21 \text{ eV}$$

2. Apply equation:

   $$KE_{max} = E - \phi = 6.21 - 4.2 = 2.01 \text{ eV}$$

Example 2: Finding Threshold Frequency

Problem: The work function of sodium is 2.75 eV. Calculate its threshold wavelength.

Solution:

$$\lambda_0 = \frac{hc}{\phi} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{2.75 \times 1.6 \times 10^{-19}}$$ $$\lambda_0 = 4.52 \times 10^{-7} \text{ m} = 452 \text{ nm}$$

This is in the blue-violet region of visible light.

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Important Points for JEE

Quick Formulas

Quantity	Formula
Work function	$\phi = h\nu_0$
Maximum KE	$KE_{max} = h\nu - \phi$
Stopping potential	$V_0 = \frac{KE_{max}}{e}$
Threshold wavelength	$\lambda_0 = \frac{hc}{\phi}$
Maximum velocity	$v_{max} = \sqrt{\frac{2(h\nu - \phi)}{m}}$

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Practice Problems

1. Light of frequency $1.5 \times 10^{15}$ Hz ejects electrons with maximum KE of 2 eV. Find the work function.

2. Two metals A and B have work functions 4 eV and 2 eV. What is the ratio of their threshold wavelengths?

3. If the intensity of light is doubled (keeping frequency same), how does the stopping potential change?

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