Dual Nature of Radiation & Matter — Photoelectric Effect

For: Akhil | Exam: JEE Main + Advanced | Subject: Modern Physics

1. The Core Idea — Wave–Particle Duality

• Light behaves as a wave (interference, diffraction, polarization) and as a particle (photoelectric effect, Compton effect).
• Matter (electrons, etc.) also shows wave nature (de Broglie waves, electron diffraction).
• Neither picture alone is complete — which nature shows up depends on the experiment.

2. The Photon

Light energy comes in discrete packets called photons.

Momentum of a photon (it has momentum but zero rest mass):

Useful working constant:

• $h = 6.63\times10^{-34}\ \text{J·s}$
• Photon travels at $c$; energy $\propto$ frequency, not intensity.
• Intensity of light = number of photons per second per unit area (each photon still carries the same $h\nu$).

3. Photoelectric Effect

When light of high enough frequency falls on a metal surface, electrons are ejected. These are photoelectrons.

Key definitions

• Work function $\phi_0$ — minimum energy to free the most loosely bound electron.
• Threshold frequency $\nu_0$ — minimum frequency for emission: $\phi_0 = h\nu_0$.
• Threshold wavelength $\lambda_0 = \dfrac{hc}{\phi_0}$.
• Stopping potential $V_0$ — reverse voltage that just stops the fastest electron.

Einstein’s Photoelectric Equation

The Four Experimental Laws

1. Emission is instantaneous (no time lag) — supports particle model.
2. Emission occurs only if $\nu \ge \nu_0$, whatever the intensity.
3. $K_{\max}$ (and $V_0$) depend only on frequency, not on intensity.
4. Photocurrent (number of electrons/sec) $\propto$ intensity (for $\nu > \nu_0$).

Why classical wave theory fails: it predicts KE should rise with intensity and allows a time lag at low intensity — both are wrong. Einstein’s photon model fixes this.

4. Important Graphs (know the slopes)

5. de Broglie Wavelength (Matter Waves)

Every moving particle has a wavelength:

For a charged particle accelerated through potential $V$:

For an electron (handy shortcut):

• Larger mass / higher speed → smaller $\lambda$ → wave nature negligible for everyday objects.
• Confirmed by the Davisson–Germer experiment (electron diffraction).

6. Common Traps (JEE loves these)

• Increasing intensity raises current, not $K_{\max}$ or $V_0$.
• Increasing frequency raises $K_{\max}$ and $V_0$, not saturation current.
• Stopping potential is a potential (volts); $eV_0$ is the energy — don’t mix units.
• No emission at all if $\nu < \nu_0$, however bright the light.
• A photon’s momentum is $h/\lambda$ even though its rest mass is zero.
• Use $E(\text{eV}) = 1240/\lambda(\text{nm})$ to save time.

7. Worked Practice Questions

Q1 (JEE Main level)

Light of wavelength $400\ \text{nm}$ falls on a metal of work function $2.0\ \text{eV}$. Find the maximum kinetic energy of the emitted electrons and the stopping potential.

Solution

Answer: $K_{\max} = 1.1\ \text{eV}$, $V_0 = 1.1\ \text{V}$.

Q2 (Threshold)

The work function of a metal is $2.48\ \text{eV}$. Find its threshold wavelength. Will light of $500\ \text{nm}$ cause emission?

Solution

Answer: $\lambda_0 = 500\ \text{nm}$; 500 nm light is exactly at threshold (borderline emission, $K_{\max}\approx 0$).

Q3 (de Broglie)

Find the de Broglie wavelength of an electron accelerated through $100\ \text{V}$.

Solution

Answer: $\lambda_e \approx 1.23\ \text{Å}$.

Q4 (Graph reasoning — Advanced flavour)

In a photoelectric experiment, the stopping potential is $1.5\ \text{V}$ for incident frequency $\nu_1$ and $3.0\ \text{V}$ for $\nu_2 = 2\nu_1$. Find the work function (in eV) in terms of $h\nu_1$.

Solution

Answer: $h\nu_1 = 1.5\ \text{eV}$ and $\phi_0 = 0$ — i.e. these numbers describe a hypothetical metal with $\nu_0 \to 0$. Lesson: always verify consistency; if $\phi_0$ comes out non-physical, the data (or your reading of it) needs a re-check.

8. Try These Yourself (answers at the end)

1. A photon has energy $4.0\ \text{eV}$. Find its wavelength (nm) and momentum.
2. Metal X has $\phi_0 = 3.0\ \text{eV}$. Light of $300\ \text{nm}$ falls on it. Find $V_0$.
3. A proton and an electron have the same kinetic energy. Which has the larger de Broglie wavelength, and why?

<details> <summary>Answers</summary>

1. $\lambda = 1240/4.0 = 310\ \text{nm}$; $p = E/c = (4.0\times1.6\times10^{-19})/(3\times10^8) = 2.13\times10^{-27}\ \text{kg·m/s}$.
2. $E = 1240/300 = 4.13\ \text{eV}$; $V_0 = 4.13 - 3.0 = 1.13\ \text{V}$.
3. Electron — same $K$ but smaller mass, and $\lambda = h/\sqrt{2mK}$, so smaller $m$ → larger $\lambda$.

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Tags

Modern Physics · Dual Nature · Photoelectric Effect · de Broglie · JEE Main + Advanced