Photoelectric Effect — The Experiment That Changed Physics

In 1887, Hertz noticed something strange: ultraviolet light hitting a metal surface caused sparks to jump more easily. In 1905, Einstein explained why — and that explanation won him the Nobel Prize and launched quantum mechanics.

The photoelectric effect — the emission of electrons when light hits a metal surface — sounds simple. But it demolished the wave theory of light and proved that light comes in discrete packets (photons). Everything quantum-mechanical flows from this insight.

In CBSE Class 12 and JEE, this is a 5–8 mark chapter. The experimental observations and Einstein’s explanation are asked almost every year.

Key Terms & Definitions

Photoelectric Effect: Emission of electrons from a metal surface when light of sufficient frequency is incident on it.

Photoelectron: An electron emitted by the photoelectric effect.

Work Function ( $\phi$ or $W$ ): The minimum energy required to eject an electron from the metal surface. Each metal has a characteristic work function. Unit: eV (electron volt) or Joule.

Threshold Frequency ( $\nu_0$ ): The minimum frequency of light required to eject electrons from a metal. Below this frequency, no electrons are emitted regardless of intensity. $\phi = h\nu_0$ .

Threshold Wavelength ( $\lambda_0$ ): Corresponding maximum wavelength: $\lambda_0 = c/\nu_0$ . Above $\lambda_0$ , no emission occurs.

Stopping Potential ( $V_0$ ): The minimum negative potential applied to the collector that stops ALL photoelectrons — even the most energetic ones. Measures maximum kinetic energy: $eV_0 = KE_{max}$ .

Photon: A quantum of electromagnetic radiation with energy $E = h\nu = hc/\lambda$ .

Einstein’s Photoelectric Equation

KE_{max} = h\nu - \phi = h(\nu - \nu_0)

Or equivalently:

eV_0 = h\nu - \phi

Where:

$h = 6.626 \times 10^{-34}$ J·s (Planck’s constant)
$\nu$ = frequency of incident light
$\phi$ = work function of metal
$V_0$ = stopping potential

This single equation explains everything the wave theory couldn’t.

The Four Experimental Observations and Why Classical Theory Failed

Observation 1: No emission below threshold frequency

Classical prediction: With enough intensity, any frequency should eventually knock out electrons (just accumulate energy).

What actually happens: Below $\nu_0$ , NO electrons are emitted, no matter how intense the light or how long it shines.

Einstein’s explanation: Each photon has energy $h\nu$ . If $h\nu < \phi$ , one photon cannot give an electron enough energy to escape. More intense light just means more photons — still none with enough energy. The interaction is one photon → one electron.

Observation 2: Instantaneous emission

Classical prediction: For dim light, electrons should need time to accumulate enough energy — significant time delay expected.

What actually happens: Emission is instantaneous (within $10^{-9}$ s), even for extremely dim light.

Einstein’s explanation: One photon transfers all its energy to one electron immediately. No accumulation needed.

Observation 3: Kinetic energy depends on frequency, not intensity

Classical prediction: More intense light = more energy = faster electrons.

What actually happens: $KE_{max}$ depends only on frequency ( $KE_{max} = h\nu - \phi$ ). Intensity determines the NUMBER of electrons, not their speed.

Einstein’s explanation: Each electron receives exactly one photon’s worth of energy. Higher frequency → higher energy photon → faster electron.

Observation 4: Saturation current proportional to intensity

Increasing light intensity increases the number of photons → more photoelectrons → larger photocurrent. Intensity does not change stopping potential.

Solved Examples

Example 1 — Easy (CBSE Level)

The work function of sodium is 2.3 eV. Find the threshold wavelength.

Solution: $\phi = h\nu_0 = hc/\lambda_0$

\lambda_0 = \frac{hc}{\phi} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{2.3 \times 1.6 \times 10^{-19}}

= \frac{19.88 \times 10^{-26}}{3.68 \times 10^{-19}} = 5.4 \times 10^{-7} \text{ m} = 540 \text{ nm}

Light of wavelength > 540 nm cannot cause photoelectric emission from sodium.

Example 2 — Medium (JEE Level)

Light of wavelength 400 nm is incident on cesium (work function = 2.0 eV). Find the maximum KE and stopping potential.

Solution:

E_{photon} = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}}

= \frac{19.88 \times 10^{-26}}{4 \times 10^{-7}} = 4.97 \times 10^{-19} \text{ J} = \frac{4.97 \times 10^{-19}}{1.6 \times 10^{-19}} \text{ eV} = 3.1 \text{ eV}

KE_{max} = 3.1 - 2.0 = \mathbf{1.1 \text{ eV}}

V_0 = 1.1 \text{ V}

Example 3 — Hard (JEE Main Style)

The stopping potential for photoelectric emission is $V_0$ when light of frequency $\nu$ is used. What is the stopping potential when light of frequency $2\nu$ is used? (Work function = $\phi$ )

Solution: For frequency $\nu$ : $eV_0 = h\nu - \phi$

For frequency $2\nu$ : $eV_0' = h(2\nu) - \phi = 2h\nu - \phi$

eV_0' = 2h\nu - \phi = (h\nu - \phi) + h\nu = eV_0 + h\nu

V_0' = V_0 + \frac{h\nu}{e}

Example 4 — Graphical Analysis

Stopping potential $V_0$ vs frequency $\nu$ is a straight line: $V_0 = \frac{h}{e}\nu - \frac{\phi}{e}$ .

Slope = $h/e$ → gives Planck’s constant (Millikan used this to measure $h$ )

X-intercept = $\nu_0$ (threshold frequency)

Y-intercept = $-\phi/e$ (negative, gives work function)

Exam-Specific Tips

CBSE 2024 asked: “Draw and explain the graph of stopping potential vs frequency.” Know that slope $= h/e$ , x-intercept $= \nu_0$ , and the line shifts vertically for different metals (different $\phi$ ) but has the same slope (since $h$ is universal).

JEE Main 2024: A question gave threshold wavelength and incident wavelength; asked for KE. Convert wavelength to energy using $E = hc/\lambda$ , then subtract work function. Always convert to consistent units (either all in eV or all in Joules).

A useful shortcut: $E \text{ (eV)} = \frac{1240}{\lambda \text{ (nm)}}$ . This comes from $hc = 1240$ eV·nm. It saves huge calculation time in MCQ exams.

Common Mistakes to Avoid

Mistake 1: Thinking higher intensity means higher KE. Intensity determines the NUMBER of photoelectrons (photocurrent), not their energy. Maximum KE depends only on the FREQUENCY of light (via $KE = h\nu - \phi$ ). This is the most tested conceptual point.

Mistake 2: Units confusion — mixing eV and Joules. Work function is often given in eV, photon energy calculated in Joules. Convert before subtracting. Conversion: $1 \text{ eV} = 1.6 \times 10^{-19}$ J.

Mistake 3: Saying emission stops at threshold frequency. It’s the opposite: threshold is the MINIMUM frequency for emission. At exactly $\nu_0$ , electrons are emitted with zero kinetic energy.

Mistake 4: Confusing stopping potential and saturation current. Stopping potential $V_0$ stops all electrons (measures $KE_{max}$ ). Saturation current is the maximum current when all emitted electrons are collected. These are different quantities on the $I$ - $V$ curve.

Mistake 5: Applying wave theory to explain observations. If an exam asks “why does classical theory fail?”, always point to the FOUR specific observations — especially the threshold frequency and instant emission. Don’t just say “wave theory is wrong.”

Practice Questions

Q1. Ultraviolet light of frequency $1.5 \times 10^{15}$ Hz hits a metal with work function 4.2 eV. Find stopping potential. ( $h = 4.14 \times 10^{-15}$ eV·s)

$E = h\nu = 4.14 \times 10^{-15} \times 1.5 \times 10^{15} = 6.21$ eV

$V_0 = KE_{max}/e = (6.21 - 4.2) = 2.01$ V

Q2. For what wavelength of light is the KE of photoelectrons from platinum (work function 5.65 eV) equal to 1 eV?

Required photon energy = $5.65 + 1 = 6.65$ eV

$\lambda = \frac{1240}{6.65} = 186$ nm (ultraviolet range)

Q3. In a photoelectric experiment, the stopping potential is 2 V. What is the maximum velocity of photoelectrons? ( $m_e = 9.1 \times 10^{-31}$ kg)

$KE_{max} = eV_0 = 1.6 \times 10^{-19} \times 2 = 3.2 \times 10^{-19}$ J

$\frac{1}{2}mv^2 = 3.2 \times 10^{-19}$

$v = \sqrt{\frac{2 \times 3.2 \times 10^{-19}}{9.1 \times 10^{-31}}} = \sqrt{7.03 \times 10^{11}} \approx 8.4 \times 10^5$ m/s

Q4. Two metals A and B have work functions 2 eV and 4 eV. Which has lower threshold frequency? Why?

Metal A has lower threshold frequency. $\nu_0 = \phi/h$ . Lower work function → lower minimum energy needed → lower threshold frequency. Metal A requires less energetic photons to start emission.

Additional Worked Examples

Example 5 — Comparing Two Metals

Two metals A and B have work functions 2.5 eV and 5.0 eV. Light of wavelength 200 nm is incident on both. Calculate the ratio of maximum velocities of photoelectrons from A and B.

E = \frac{1240}{200} = 6.2 \text{ eV}

$KE_A = 6.2 - 2.5 = 3.7$ eV

$KE_B = 6.2 - 5.0 = 1.2$ eV

Since $KE = \frac{1}{2}mv^2$ and the electron mass is the same:

\frac{v_A}{v_B} = \sqrt{\frac{KE_A}{KE_B}} = \sqrt{\frac{3.7}{1.2}} = \sqrt{3.083} \approx 1.76

Example 6 — Power and Number of Electrons

A 100 W sodium lamp emits light of wavelength 589 nm. If 5% of the light energy falls on a metal surface (work function 1.9 eV) and all incident photons eject electrons, find the number of photoelectrons emitted per second.

E = \frac{1240}{589} = 2.11 \text{ eV} = 2.11 \times 1.6 \times 10^{-19} = 3.37 \times 10^{-19} \text{ J}

$P_{surface} = 0.05 \times 100 = 5$ W

n = \frac{P}{E} = \frac{5}{3.37 \times 10^{-19}} = 1.48 \times 10^{19} \text{ photons/s}

Since every photon ejects one electron: $1.48 \times 10^{19}$ photoelectrons per second.

I_{saturation} = n_e \times e

where $n_e$ = number of photoelectrons emitted per second, $e = 1.6 \times 10^{-19}$ C.

The saturation current is reached when all emitted electrons are collected (no recombination). Increasing intensity increases $n_e$ (and hence $I_{sat}$ ) but does NOT change $KE_{max}$ or $V_0$ .

NEET 2022 asked about the relationship between photocurrent and intensity. The answer: photocurrent is directly proportional to intensity (since intensity determines the number of photons per second, and each photon ejects at most one electron). Stopping potential is independent of intensity.

Q5. Light of wavelength 300 nm falls on a metal with work function 2.13 eV. Find the de Broglie wavelength of the fastest photoelectrons emitted.

$E_{photon} = 1240/300 = 4.13$ eV. $KE_{max} = 4.13 - 2.13 = 2.0$ eV $= 3.2 \times 10^{-19}$ J.

$\lambda_{dB} = \frac{h}{\sqrt{2mKE}} = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 3.2 \times 10^{-19}}}$

$= \frac{6.626 \times 10^{-34}}{7.63 \times 10^{-25}} = 8.68 \times 10^{-10}$ m $\approx$ 8.7 Angstrom

FAQs

What did the photoelectric effect prove about light?

It proved that light is quantised — it comes in discrete packets (photons) of energy $E = h\nu$ . This was the first experimental proof that contradicted the continuous wave model of light. It showed light has particle-like properties in its interactions with matter.

Why does the photoelectric effect not work for all metals with visible light?

Metals with high work functions (like platinum, tungsten) require UV light to provide sufficient photon energy. Alkali metals (sodium, potassium, cesium) have lower work functions and respond to visible light. The specific threshold depends on the metal’s electronic structure.

What is the difference between photoelectric effect and fluorescence?

In the photoelectric effect, absorbed photon energy ejects an electron. In fluorescence, absorbed photon energy excites an electron to a higher orbital, and when it falls back, it emits a photon of lower energy (longer wavelength). Both involve photon absorption, but the outcome differs.

Can the photoelectric effect produce X-rays?

No — the photoelectric effect uses X-rays as input to eject electrons. The reverse process (electrons hitting a metal and producing X-rays) is called Bremsstrahlung or characteristic radiation.

What is the significance of Planck’s constant in this context?

Planck’s constant $h$ is the proportionality factor between frequency and photon energy. Millikan measured $h$ precisely by plotting $V_0$ vs $\nu$ — the slope gives $h/e$ . His value matched Planck’s independently derived constant, confirming the quantum picture.