KE = hν - φ. Below threshold, no electrons ejected regardless of intensity.

Photoelectric Effect — Threshold Frequency and Work Function

Question

A metal surface has a work function of 2.4 eV. Light of frequency $7.0 \times 10^{14}$ Hz falls on it. Find:

The threshold frequency of the metal
The maximum kinetic energy of the emitted photoelectrons

(Given: $h = 6.626 \times 10^{-34}$ J·s, $1 \text{ eV} = 1.6 \times 10^{-19}$ J)

Solution — Step by Step

The work function $\phi = 2.4 \text{ eV}$ needs to be in Joules to match our units for $h$ .

\phi = 2.4 \times 1.6 \times 10^{-19} = 3.84 \times 10^{-19} \text{ J}

Threshold frequency is the minimum frequency at which ejection just begins — meaning all the photon energy goes into overcoming the work function, with nothing left over as KE.

\nu_0 = \frac{\phi}{h} = \frac{3.84 \times 10^{-19}}{6.626 \times 10^{-34}}

\nu_0 = 5.79 \times 10^{14} \text{ Hz}

Since the incident frequency $\nu = 7.0 \times 10^{14}$ Hz is greater than $\nu_0 = 5.79 \times 10^{14}$ Hz, photoelectric effect does occur. If $\nu < \nu_0$ , we stop here — no electrons come out, full stop.

KE_{max} = h\nu - \phi

KE_{max} = (6.626 \times 10^{-34})(7.0 \times 10^{14}) - 3.84 \times 10^{-19}

KE_{max} = 4.638 \times 10^{-19} - 3.84 \times 10^{-19}

KE_{max} = 7.98 \times 10^{-20} \text{ J}

KE_{max} = \frac{7.98 \times 10^{-20}}{1.6 \times 10^{-19}} \approx \textbf{0.5 eV}

Why This Works

Einstein treated light as a stream of photons, each carrying energy $E = h\nu$ . When one photon hits one electron, it’s an all-or-nothing transaction — the electron uses $\phi$ joules to escape the metal surface, and whatever energy is left over becomes kinetic energy. This is why the equation is simply subtraction: $KE = h\nu - \phi$ .

The threshold frequency exists because a photon with $\nu < \nu_0$ simply doesn’t have enough energy to pay the “exit fee” ( $\phi$ ). No matter how many such photons you throw at the surface — no matter how intense the light — no single photon can give an electron enough energy. This is what shattered the classical wave theory of light, which predicted that enough intensity should always work.

This experiment directly gives us experimental proof for the quantum nature of light. The stopping potential in JEE problems is just $KE_{max}$ expressed in eV — they’re the same number.

Alternative Method

In JEE MCQs, work entirely in eV to save calculation time.

Convert the incident photon energy directly to eV:

E = h\nu = (6.626 \times 10^{-34})(7.0 \times 10^{14}) = 4.638 \times 10^{-19} \text{ J}

E = \frac{4.638 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.9 \text{ eV}

Now simply subtract the work function:

KE_{max} = 2.9 - 2.4 = \textbf{0.5 eV}

Much faster. In JEE Main 2023, the numbers were set up so that working in eV saves roughly 40 seconds — meaningful under exam conditions.

Common Mistake

Confusing intensity with frequency. Students often think: “if we double the intensity, electrons will come out even below threshold.” Wrong. Intensity only increases the number of photons per second, not the energy of each photon. A higher-intensity beam of sub-threshold light ejects zero electrons — same as a low-intensity beam. Only frequency determines whether ejection happens. Once above threshold, intensity increases the number of emitted electrons (photoelectric current), not their maximum KE.

A related trap in numericals: if a question gives intensity in W/m² instead of frequency, you must first find the energy per photon from $E = h\nu$ , then figure out how many photons per second hit the surface. The intensity route won’t give you KE directly.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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