Modern Physics: Speed-Solving Techniques (4)

Question

Light of wavelength $400$ nm is incident on a metal of work function $2.0$ eV. Find (a) the maximum kinetic energy of the photoelectrons, and (b) the stopping potential.

Solution — Step by Step

Final answers: (a) $KE_{\max} = 1.1$ eV, (b) $V_0 = 1.1$ V.

Why This Works

Each photon carries energy $h\nu = hc/\lambda$. When the photon hits the metal, it gives all its energy to one electron. The electron uses up the work function $\phi$ to escape the surface, and the leftover energy becomes kinetic energy.

The “1240 eV·nm” constant is just $hc$ in convenient units. Memorise it — it saves you from computing $h\nu$ in joules and converting back.

Alternative Method

Compute in SI units. $E = hc/\lambda = (6.63\times 10^{-34})(3\times 10^8)/(400\times 10^{-9}) = 4.97\times 10^{-19}$ J. Divide by $1.6\times 10^{-19}$ to get $3.1$ eV. Slow, but the SI route is what the textbook shows.

Common Mistake

Mixing units — using $\lambda$ in metres but expecting an answer in eV. The shortcut $1240/\lambda$ requires $\lambda$ in nanometres. If $\lambda$ is given in angstroms, convert to nm first (divide by 10).