E.2.3 (HL)—Photoelectric equation

Track the energy budget

One photon transfers energy $hf$ to one electron. The work function is spent releasing the electron; any remainder is its maximum kinetic energy. Use joules or electronvolts consistently.

E_{k,\max}=hf-\Phi=\frac{hc}{\lambda}-\Phi

Worked example — 420 nm light

For $\lambda=420\,\mathrm{nm}$ , photon energy is $hc/\lambda=2.96\,\mathrm{eV}$ . With $\Phi=2.0\,\mathrm{eV}$ , $E_{k,\max}=2.96-2.0=0.96\,\mathrm{eV}$ . The result is positive, so emission occurs; a negative calculated remainder would mean no emission.

Use wavelength when given

Because $f=c/\lambda$ , write $E_{k,max}=\frac{hc}{\lambda}-\Phi$ . Keep $hc/\lambda$ and $\Phi$ in the same energy unit before subtracting.

Find maximum speed

Once $E_{k,max}$ is known, use $E_{k,max}=\frac12m_ev_{max}^2$ , so $v_{max}=\sqrt{2E_{k,max}/m_e}$ .

Common trap

Do not add the work function to the kinetic energy. The work function is the energy already spent escaping the surface.

How it is assessed

Questions calculate work function from wavelength and kinetic energy or derive maximum speed from photon energy and work function.

Command terms

Calculate / Determine

What earns marks

Use hc/lambda or hf, subtract Phi once, then convert the remaining kinetic energy to speed only after unit consistency is established.

Watch for

Using the wrong sign for Phi, mixing joules and electron-volts, or omitting the factor 2 in the kinetic-energy speed relation.

Retrieve the light model

The photoelectric effect and Compton scattering show photon-like energy and momentum transfer. Threshold frequency and $E_{k,max}=hf-\Phi$ make the photon energy budget explicit.

Retrieve the matter model

Particle diffraction and $\lambda=h/p$ show wave-like matter. For Compton scattering, track energy loss, increased wavelength, and $\Delta\lambda=\frac{h}{m_ec}(1-\cos\theta)$ .

E.2.3 (HL)—Photoelectric equation

Use the Photoelectric Equation

E.2.3 (HL) Exam Analysis

Assessment in practice

Retrieve the Quantum Model