Question
The work function of a metal is ϕ=2.5 eV. Light of wavelength λ=400 nm falls on it. Find (a) the maximum kinetic energy of photoelectrons, (b) the stopping potential, (c) the maximum wavelength that can produce photoelectrons (threshold wavelength). Take h=6.63×10−34 J s, c=3×108 m/s, 1 eV=1.6×10−19 J.
Solution — Step by Step
E=hc/λ=400×10−96.63×10−34×3×108=4.97×10−19 J
In eV: E=4.97×10−19/(1.6×10−19)=3.11 eV.
(Quick shortcut: E(eV)=1240/λ(nm)=1240/400=3.1 eV.)
KEmax=E−ϕ=3.1−2.5=0.6 eV=0.6×1.6×10−19=9.6×10−20 J.
eVs=KEmax, so Vs=KEmax/e=0.6 V.
At threshold, hc/λ0=ϕ. So λ0(nm)=1240/ϕ(eV)=1240/2.5=496 nm.
Photons with λ>496 nm cannot produce photoelectrons, no matter how intense the light.
Final answers: KEmax=0.6 eV, Vs=0.6 V, λ0=496 nm.
Why This Works
The photoelectric effect’s signature feature: energy per photon depends on frequency, not intensity. Below threshold (ν<ν0, equivalently λ>λ0), no photoelectrons emerge no matter how bright the light — each photon individually lacks enough energy to free an electron.
The shortcut E(eV)=1240/λ(nm) is gold for JEE/NEET. Memorize it. It comes from hc=1240 eV nm.
Alternative Method
Use λ0=hc/ϕ directly to get threshold first, then KEmax=(hc/λ)(1−λ/λ0)=3.1(1−400/496)=3.1×0.193=0.6 eV. Same answer, cleaner if you have λ0 already.
Common Mistake
A frequent NEET trap: students mix up frequency and wavelength conditions. “Threshold frequency” ν0 is the minimum frequency for emission, but “threshold wavelength” λ0 is the maximum wavelength. Below λ0 — emission happens; above λ0 — no emission. Always cross-check with E=hc/λ.