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Most photoelectric questions in JEE Main and your board exam turn on one comparison: is the energy of one photon bigger than the metal's work function? Turn that comparison into a wavelength and you get the threshold wavelength, the longest wavelength that still knocks electrons out. This post shows the one-line formula, where the number 1240 comes from, and why a red laser pointer does nothing to a zinc plate.
What is the threshold wavelength?
The work function φ is the least energy needed to pull one electron out of a metal's surface. A photon of wavelength λ carries energy E = hc/λ. Electrons come out only when E is at least φ, so the boundary sits where hc/λ0 = φ.
Threshold wavelength: λ0 = hc/φ. Any wavelength longer than λ0 fails; any shorter one works, and the leftover energy becomes the electron's kinetic energy, KEmax = hc/λ − φ.
Notice the direction. A bigger work function means a shorter threshold wavelength, because a tightly bound electron needs a higher-energy photon.
Where does 1240 eV nm come from?
The three constants you need are fixed exactly in the SI. NIST lists the Planck constant as exactly 6.626 070 15 × 10⁻³⁴ J s (NIST, Planck constant), the speed of light as exactly 299 792 458 m/s (NIST, speed of light), and the elementary charge as exactly 1.602 176 634 × 10⁻¹⁹ C (NIST, elementary charge).
Multiply h by c and divide by e to change joules into electronvolts: hc = 1.2398 × 10⁻⁶ eV m, which is 1239.8 eV nm. Round it to 1240 and the formula becomes:
λ0 (nm) = 1240 / φ (eV).
That saves you from writing 10⁻³⁴ and 10⁻¹⁹ in the exam hall. If you want the same shortcut for photon energy, our post on the E = 1240/λ photon-energy formula works through it in the other direction.
Why does red light fail to eject electrons from some metals?
NASA gives the visible range as roughly 380 to 700 nanometres, with red at the long end near 700 nm (NASA, visible light). One photon of 700 nm light carries 1240/700 = 1.77 eV.
Now compare that with work functions. The table on HyperPhysics, taken from the CRC Handbook of Chemistry and Physics, lists about 2.1 eV for cesium, 2.28 eV for sodium, 4.3 eV for zinc, 4.7 eV for copper and 6.35 eV for platinum (HyperPhysics, work functions). Each of these is more than 1.77 eV, so a red photon cannot free an electron from any of them.
Brightness does not rescue it. A brighter beam sends more photons per second, but each electron still absorbs one photon at a time, and each red photon is too weak. That is why intensity changes the photocurrent but never the threshold.
Run the same table through λ0 = 1240/φ:
Cesium, 2.1 eV: λ0 ≈ 590 nm, so violet, blue, green and yellow light work, but orange and red do not.
Sodium, 2.28 eV: λ0 ≈ 544 nm, so violet, blue and bluish-green light work, but yellow, orange and red do not.
Zinc, 4.3 eV: λ0 ≈ 288 nm, in the ultraviolet, so no visible colour works at all.
Platinum, 6.35 eV: λ0 ≈ 195 nm, deep in the ultraviolet.
The hero diagram above plots three of these metals: the photon-energy curve crosses each metal's work-function line at its threshold wavelength. For more on which metals respond to visible light, read which metals show the photoelectric effect with visible light.
How do you solve a threshold wavelength question step by step?
Use this four-step routine on any problem.
Write φ in eV. Convert from joules if needed.
Find λ0 = 1240/φ in nm.
Compare the given wavelength with λ0. Longer means no emission, and the answer stops here.
If it is shorter, find KEmax = 1240/λ − φ in eV, and the stopping potential in volts has the same number.
Worked example, using sodium's 2.28 eV from the HyperPhysics table. The threshold is 1240/2.28 ≈ 544 nm. Light of 700 nm is longer, so no electrons come out. Light of 400 nm is shorter: each photon carries 1240/400 = 3.10 eV, so KEmax = 3.10 − 2.28 = 0.82 eV and the stopping potential is 0.82 V.
Examiners also run this backwards, giving two wavelengths and two stopping potentials and asking for φ and h. Our walk-through on finding the work function from two wavelengths covers that version.
If you keep getting the unit conversion or the direction of the inequality wrong, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer, so you can see which step slipped. For ideas that are easier to see than read, the Fermi Simulation Lab has interactive simulations.
