Question
The work function of a metal is ϕ=2.5 eV. Light of wavelength λ=400 nm is incident on it. Find (a) the maximum kinetic energy of emitted photoelectrons, (b) the stopping potential, and (c) the threshold wavelength of the metal. (Take h=6.626×10−34 J·s, c=3×108 m/s, 1 eV =1.6×10−19 J.)
Solution — Step by Step
E=λhc=400×10−96.626×10−34×3×108≈4.97×10−19 J
In eV: E=4.97×10−19/(1.6×10−19)≈3.10 eV.
KEmax=E−ϕ=3.10−2.5=0.60 eV.
Vs=KEmax/e=0.60 V (since KE is in eV, dividing by e in coulombs and multiplying by 1 V gives the number directly).
At threshold, E=ϕ, so λ0=hc/ϕ:
λ0=2.5 eV1240 eV⋅nm=496 nm
Final answers: KEmax=0.60 eV, Vs=0.60 V, λ0≈496 nm.
Why This Works
Einstein’s photoelectric equation KEmax=hν−ϕ is the entire chapter in one line. Every PYQ on photoelectric effect reduces to plugging into this. The shortcut hc≈1240 eV·nm makes the arithmetic 30 seconds faster.
The threshold wavelength is the longest λ that can still kick an electron out — at that point all the photon energy goes into work function and the electron just barely escapes with zero KE.
Alternative Method
Convert λ to frequency first: ν=c/λ=7.5×1014 Hz. Then E=hν=4.97×10−19 J. Same answer through a longer path.
Memorise hc=1240 eV·nm. With λ in nm and ϕ in eV, the photoelectric problem becomes pure arithmetic in seconds.
Common Mistake
Using λ>λ0 in the photoelectric equation and getting negative KE. If λ>λ0 (or equivalently E<ϕ), no emission happens — KEmax=0, not negative. The equation only applies when E≥ϕ.
Don’t forget the maximum in KEmax. Photoelectrons come out with a range of energies between 0 and KEmax. Stopping potential corresponds to KEmax, not the average.