Photoelectric effect — threshold frequency, stopping potential, Einstein's equation

Question

Light of wavelength $400$ nm falls on a metal with work function $\phi = 2.0$ eV. Find (a) the maximum kinetic energy of photoelectrons and (b) the stopping potential. Will emission occur if the wavelength is changed to $700$ nm?

(JEE Main & NEET — tested every year)

Solution — Step by Step

E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} = 4.97 \times 10^{-19} \text{ J}

Converting to eV: $E = \dfrac{4.97 \times 10^{-19}}{1.6 \times 10^{-19}} = 3.1$ eV

Shortcut: use $E(\text{eV}) = \dfrac{1240}{\lambda(\text{nm})} = \dfrac{1240}{400} = 3.1$ eV

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

This is the maximum KE because we assume the photon energy goes entirely to ejecting the electron with maximum speed (no energy lost inside the metal).

The stopping potential $V_0$ is the voltage needed to stop the fastest photoelectrons:

eV_0 = KE_{\max}

V_0 = \frac{KE_{\max}}{e} = \frac{1.1 \text{ eV}}{e} = \mathbf{1.1 \text{ V}}

When KE is in eV, the stopping potential in volts has the same numerical value.

E = \frac{1240}{700} = 1.77 \text{ eV}

Since $1.77 < \phi = 2.0$ eV, the photon energy is less than the work function. No photoelectric emission occurs, regardless of light intensity.

Why This Works

Einstein’s equation is energy conservation at the photon level: one photon gives all its energy to one electron. Part of the energy ( $\phi$ ) goes to overcome the binding, and the rest becomes kinetic energy.

graph TD
    A["Light hits metal"] --> B{"E = hf vs φ?"}
    B -->|"hf < φ"| C["No emission<br/>(even with high intensity)"]
    B -->|"hf ≥ φ"| D["Emission occurs"]
    D --> E["KE_max = hf - φ"]
    E --> F["Stopping potential<br/>V₀ = KE_max / e"]
    D --> G["Key observations"]
    G --> G1["KE_max depends on<br/>frequency, NOT intensity"]
    G --> G2["Current depends on<br/>intensity (more photons)"]
    G --> G3["Emission is<br/>instantaneous"]

The wave theory of light fails here — it predicts that brighter light should always cause emission (given enough time) and that KE should depend on intensity. Neither is true. This was the evidence that convinced physicists light has particle nature.

Alternative Method — Graphical Analysis

Plot $KE_{\max}$ vs frequency $f$ : you get a straight line with slope $h$ and x-intercept at the threshold frequency $f_0 = \phi/h$ . This graph is a JEE favourite. The slope is the same for all metals (it’s Planck’s constant), but the x-intercept varies with the metal (different work functions).

The shortcut $E = 1240/\lambda$ (with $\lambda$ in nm and $E$ in eV) saves enormous time. Memorise it — you’ll use it in every photoelectric effect and energy-level problem. It comes from $hc = 1240$ eV nm.

Common Mistake

Students think increasing intensity below the threshold frequency will eventually cause emission if you wait long enough. This is wrong — no matter how intense the light, if each photon has energy less than $\phi$ , no electron can escape. A million low-energy photons cannot combine their energy to free one electron (each photon interacts independently). This is the core concept that NEET tests repeatedly.

Question

Solution — Step by Step

Why This Works

Alternative Method — Graphical Analysis

Common Mistake

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