Photoelectric EffectJEE Main

Concept

Shine light on a metal and electrons may be ejected — but only if each photon carries more energy than the metal's work function φ. Brighter light below the threshold does nothing; the effect depends on frequency, not intensity. This is Einstein's quantum picture of light.

Key formula

Derivation

Each photon delivers energy $h\nu$ to one electron. Freeing it costs $\phi$; the rest becomes kinetic energy: $K_{max} = h\nu - \phi$.

Emission needs $h\nu \ge \phi$, i.e. $\lambda \le \lambda_0 = hc/\phi$. The stopping potential $V_0$ that just halts the fastest electron satisfies $eV_0 = K_{max}$, so a plot of $V_0$ versus $\nu$ is a straight line of slope $h/e$.

Scenarios to explore

• Photoelectric Effect — Stopping potential, work function & threshold.

Real-world applications

• Photocells and light meters.
• Solar panels (photovoltaic effect).
• Night-vision photomultipliers and CCD sensors.

JEE exam tips

• Slope of the $V_0$–$\nu$ graph is $h/e$ for every metal; only the intercept changes.
• $\lambda_0(\text{nm}) = 1240/\phi(\text{eV})$ is the fastest way to the threshold.

Common mistakes

• Thinking brighter light gives faster electrons — it gives more electrons, same $K_{max}$.
• Forgetting there is no emission at all below the threshold frequency.
• Mixing eV and joules — keep everything in eV with $hc = 1240$ eV·nm.

Exam traps to avoid

• Increasing intensity raises the saturation current, not the stopping potential.
• Emission is essentially instantaneous — there is no time lag, contradicting wave theory.