Photoelectric Effect Graphs

Eight graphs that appear in NSC Physics exams — each one is just the photoelectric equation rearranged into a straight line. Every step of the algebra is shown, and each graph's slope and intercepts are explained.

Optical Phenomena
The master equation
\(E = W_0 + E_{k(max)}\)
\(E\)energy of incident photon (J), \(E = hf\)
\(W_0\)work function of the metal (J), \(W_0 = hf_0\)
\(E_{k(max)}\)max kinetic energy of photoelectron (J)
\(h\)Planck's constant (J·s)
\(c\)speed of light (m·s⁻¹), used via \(f = c/\lambda\)
\(m_e\)mass of an electron (kg), used via \(E_{k(max)} = \tfrac{1}{2}m_e v^2\)
01

\(E_{k(max)}\) vs \(f\) — Kinetic energy vs frequency

positive slope · negative y-intercept
\(f\text{ (Hz)}\)
\(E_{k(max)}\text{ (J)}\)
\(O\)
\(-W_0\)
\(f_0\)
\(m = h\)
1
Start with the photoelectric equation
\[E = W_0 + E_{k(max)}\]
2
Make Ekmax the subject
\[E_{k(max)} = E - W_0\]
3
Photon energy depends on frequency — substitute E = hf
\[E_{k(max)} = hf - W_0\]
4
Rewrite to match y = mx + c exactly
\[E_{k(max)} = h \cdot f + (-W_0)\]
5
Identify each part
\[y \to E_{k(max)},\quad x \to f,\quad m \to h,\quad c \to {-W_0}\]
Compare to y = mx + c
y=m·x+c
\(E_{k(max)}\)= \(h\)· \(f\)+ \(-W_0\)
Reading the graph
  • Slope \(= h\) — Planck's constant, the same for every metal
  • y-intercept \(= -W_0\) — negative; only the line beyond the x-intercept is physical
  • x-intercept \(= f_0\) — the threshold frequency, where \(E_{k(max)} = 0\)
  • Below \(f_0\) no electrons are ejected at all, regardless of intensity
02

\(E_{k(max)}\) vs \(\tfrac{1}{\lambda}\) — Kinetic energy vs reciprocal wavelength

positive slope · negative y-intercept
\(\tfrac{1}{\lambda}\text{ (m}^{-1}\text{)}\)
\(E_{k(max)}\text{ (J)}\)
\(O\)
\(-W_0\)
\(\tfrac{1}{\lambda_0}\)
\(m = hc\)
1
Start with the photoelectric equation, make Ekmax the subject
\[E_{k(max)} = E - W_0\]
2
Frequency relates to wavelength: f = c/λ — substitute into E = hf
\[E = hf = \frac{hc}{\lambda}\]
3
Substitute into the rearranged equation
\[E_{k(max)} = \frac{hc}{\lambda} - W_0\]
4
Rewrite 1/λ explicitly to match y = mx + c
\[E_{k(max)} = hc \cdot \frac{1}{\lambda} + (-W_0)\]
5
Identify each part
\[y \to E_{k(max)},\quad x \to \tfrac{1}{\lambda},\quad m \to hc,\quad c \to {-W_0}\]
Compare to y = mx + c
y=m·x+c
\(E_{k(max)}\)= \(hc\)· \(\tfrac{1}{\lambda}\)+ \(-W_0\)
Reading the graph
  • Slope \(= hc\) — Planck's constant × the speed of light
  • y-intercept \(= -W_0\) — same work function as in graph 01
  • x-intercept \(= \tfrac{1}{\lambda_0}\) — reciprocal of the threshold wavelength
  • Shorter wavelength (larger \(\tfrac{1}{\lambda}\)) → more energetic photoelectrons
03

\(E\) vs \(f\) — Photon energy vs frequency

straight line through origin
\(f\text{ (Hz)}\)
\(E\text{ (J)}\)
\(O\)
\(m = h\)
1
The energy of a single photon
\[E = hf\]
2
Already in y = mx + c form (c = 0)
\[E = h \cdot f + 0\]
3
Identify each part
\[y \to E,\quad x \to f,\quad m \to h,\quad c \to 0\]
Compare to y = mx + c
y=m·x+c
\(E\)= \(h\)· \(f\)+ \(0\)
Reading the graph
  • Slope \(= h\) — Planck's constant
  • Passes through the origin — a photon of zero frequency carries zero energy
  • Unlike graph 01, this plots the total photon energy, not the leftover kinetic energy — so there is no \(-W_0\) intercept
04

\(E\) vs \(\tfrac{1}{\lambda}\) — Photon energy vs reciprocal wavelength

straight line through origin
\(\tfrac{1}{\lambda}\text{ (m}^{-1}\text{)}\)
\(E\text{ (J)}\)
\(O\)
\(m = hc\)
1
Start with the photon energy equation
\[E = hf\]
2
Frequency relates to wavelength: f = c/λ — substitute
\[E = h \cdot \frac{c}{\lambda}\]
3
Rewrite 1/λ explicitly to match y = mx + c
\[E = hc \cdot \frac{1}{\lambda} + 0\]
4
Identify each part
\[y \to E,\quad x \to \tfrac{1}{\lambda},\quad m \to hc,\quad c \to 0\]
Compare to y = mx + c
y=m·x+c
\(E\)= \(hc\)· \(\tfrac{1}{\lambda}\)+ \(0\)
Reading the graph
  • Slope \(= hc\) — Planck's constant × the speed of light
  • Passes through the origin, same reasoning as graph 03
  • A longer wavelength gives a smaller \(\tfrac{1}{\lambda}\) — the point on the graph moves left, towards lower energy
05

\(v^2\) vs \(f\) — Electron speed squared vs frequency

positive slope · negative y-intercept
\(f\text{ (Hz)}\)
\(v^2\text{ (m}^2\text{s}^{-2}\text{)}\)
\(O\)
\(-\tfrac{2W_0}{m_e}\)
\(f_0\)
\(m = \tfrac{2h}{m_e}\)
1
Two expressions both equal Ekmax — set them equal
\[\tfrac{1}{2}m_e v^2 = hf - W_0\]
2
Multiply both sides by 2/me to make v² the subject
\[v^2 = \frac{2h}{m_e} \cdot f - \frac{2W_0}{m_e}\]
3
Identify each part
\[y \to v^2,\quad x \to f,\quad m \to \tfrac{2h}{m_e},\quad c \to {-\tfrac{2W_0}{m_e}}\]
Compare to y = mx + c
y=m·x+c
\(v^2\)= \(\tfrac{2h}{m_e}\)· \(f\)+ \(-\tfrac{2W_0}{m_e}\)
Reading the graph
  • Slope \(= \tfrac{2h}{m_e}\) — twice Planck's constant divided by the electron mass
  • y-intercept \(= -\tfrac{2W_0}{m_e}\) — multiply by \(-m_e/2\) to recover \(W_0\)
  • x-intercept \(= f_0\), the same threshold frequency as graph 01
  • \(v\) vs \(f\) on its own would be a curve (a square root); squaring first is what makes it linear
06

\(W_0\) vs \(f_0\) — Work function vs threshold frequency (different metals)

straight line through origin
\(f_0\text{ (Hz)}\)
\(W_0\text{ (J)}\)
\(O\)
\(m = h\)
metal A
metal D
1
Work function and threshold frequency are linked by
\[W_0 = hf_0\]
2
Already linear — each metal contributes one (f₀, W₀) point
\[W_0 = h \cdot f_0 + 0\]
3
Identify each part
\[y \to W_0,\quad x \to f_0,\quad m \to h,\quad c \to 0\]
Compare to y = mx + c
y=m·x+c
\(W_0\)= \(h\)· \(f_0\)+ \(0\)
Reading the graph
  • Slope \(= h\) — every metal's point lies on the same line because \(h\) is universal
  • Passes through the origin — a metal needing zero energy to release an electron would have zero threshold frequency
  • A metal further up the line (larger \(f_0\)) has a larger work function
07

\(V_s\) vs \(f\) — Stopping potential vs frequency

positive slope · negative y-intercept
\(f\text{ (Hz)}\)
\(V_s\text{ (V)}\)
\(O\)
\(-\tfrac{W_0}{e}\)
\(f_0\)
\(m = \tfrac{h}{e}\)
1
Start with the photoelectric equation, substitute E = hf
\[E_{k(max)} = hf - W_0\]
2
The stopping potential does work eVs to stop the fastest electron
\[eV_s = E_{k(max)}\]
3
Substitute
\[eV_s = hf - W_0\]
4
Divide both sides by e to make Vs the subject
\[V_s = \frac{h}{e}f - \frac{W_0}{e}\]
5
Identify each part
\[y \to V_s,\quad x \to f,\quad m \to \tfrac{h}{e},\quad c \to {-\tfrac{W_0}{e}}\]
Compare to y = mx + c
y=m·x+c
\(V_s\)= \(\tfrac{h}{e}\)· \(f\)+ \(-\tfrac{W_0}{e}\)
Reading the graph
  • Slope \(= \tfrac{h}{e}\) — multiply by the electron charge \(e\) to find \(h\) experimentally
  • y-intercept \(= -\tfrac{W_0}{e}\) — multiply by \(e\) to find the work function
  • x-intercept \(= f_0\) — the same threshold frequency as graph 01
  • \(V_s\) is measured directly in the lab by increasing a reverse voltage until the photocurrent stops
08

\(E_{k(max)}\) vs \(f\) — Comparing two metals

parallel lines · same slope
\(f\text{ (Hz)}\)
\(E_{k(max)}\text{ (J)}\)
\(O\)
\(-W_{0A}\)
\(f_{0A}\)
Metal A
\(-W_{0B}\)
\(f_{0B}\)
Metal B
same slope \(= h\) for both
1
The same equation applies to any metal
\[E_{k(max)} = h \cdot f + (-W_0)\]
2
Planck's constant h is a universal constant — it cannot change between metals
\[m_A = m_B = h\]
3
Only the work function W0 differs between metals
\[c_A = -W_{0A}, \quad c_B = -W_{0B}\]
4
A larger work function shifts the line down and its threshold frequency up
\[W_{0B} > W_{0A} \implies f_{0B} > f_{0A}\]
Compare to y = mx + c
Metalmc
A→ \(h\) \(-W_{0A}\)
B→ \(h\) \(-W_{0B}\)
Reading the graph
  • Both lines are parallel — equal slope confirms \(h\) is the same constant for every metal
  • The line that is lower / further right belongs to the metal with the larger work function
  • Comparing several metals this way is how Millikan experimentally confirmed Einstein's photoelectric equation and measured \(h\)