The photoelectric effect: Work function ϕ, stopping potential.

The idea

The work function ϕ is the minimum energy needed to remove an electron from the surface of a metal. A photon of frequency f supplies energy hf; the surplus becomes the maximum kinetic energy of the emitted electron, so hf = ϕ + Ek(max). The stopping potential Vs is the reverse potential difference that just prevents the fastest emitted electrons from reaching the collector, making the photocurrent zero. At that point the electrical work done on the electron equals its maximum kinetic energy: eVs = Ek(max). Combining the two gives eVs = hf − ϕ, so a graph of Vs against f is a straight line of gradient h ÷ e and intercept −ϕ ÷ e on the Vs axis. This allows ϕ and h to be found experimentally.

What earns the marks

• Work function ϕ is the minimum energy required to remove an electron from the surface of a metal, and it is a property of that metal.
• The photoelectric equation hf = ϕ + Ek(max) expresses conservation of energy for one photon absorbed by one electron.
• Ek(max) is the kinetic energy of the fastest emitted electrons; slower electrons have lost more energy before escaping.
• The stopping potential Vs is the reverse potential difference that just reduces the photocurrent to zero.
• At the stopping potential, eVs = Ek(max), so the stopping potential measures the maximum kinetic energy of the emitted electrons.
• Combining the equations gives eVs = hf − ϕ, so a graph of Vs against f has gradient h ÷ e and intercept −ϕ ÷ e on the Vs axis.
• The stopping potential depends on the frequency of the incident radiation and the metal used, but not on the intensity of the radiation.

Common mistakes and how to put them right

• Confusing the work function with the stopping potential.

  Put it right by stating that ϕ is an energy in joules (or electronvolts) while Vs is a potential difference in volts.

• Using eVs = hf without subtracting ϕ.

  Put it right by writing eVs = hf − ϕ so the energy needed to escape the metal is accounted for.

• Thinking that increasing intensity increases the stopping potential.

  Put it right by explaining that intensity changes the number of photons per second, not the energy of each photon, so Ek(max) and Vs are unchanged.

Exam tips

• Quote the photoelectric equation as hf = ϕ + Ek(max) and then substitute eVs for Ek(max) when a stopping potential is involved.
• When reading a graph of Vs against f, identify the gradient as h ÷ e and the intercept as −ϕ ÷ e before calculating.
• Keep energy units consistent: convert between joules and electronvolts using 1 eV = 1.60 × 10⁻¹⁹ J where needed.