22.2.4—Hf = Φ + 1/2 mvmax 2

One photon supplies hf. The maximum-energy electrons spend Φ escaping and keep the remainder Kmax; a stopping potential Vs removes that maximum kinetic energy because Kmax=eVs.

For f=1.10×10¹⁵ Hz and Φ=5.80×10⁻¹⁹ J: Kmax=hf−Φ=1.49×10⁻¹⁹ J, so vmax=√(2Kmax/me)=5.72×10⁵ m s⁻¹ and Vs=Kmax/e=0.933 V.

Different metals give parallel Kmax–f lines because h is universal. A larger work function shifts the threshold to a larger frequency and the extrapolated intercept to a more negative value.

Use maximum—not average—kinetic energy. If Φ is in eV, either keep the whole energy calculation in eV or convert consistently; do not write Kmax=Vs without the charge factor when using joules.