Photoelectric Calculations

Spec mapping: OCR H556 Module 4.5 — Quantum physics (numerical photoelectric problems; stopping potential $V_s = KE_{\text{max}}/e$Vs​=KEmax​/e; $V_s$Vs​-vs-$f$f graph with gradient $h/e$h/e and intercept $-\phi/e$−ϕ/e; PAG 5 anchor — Planck constant determination). Refer to the official OCR H556 specification document for exact wording.

In lesson 4 you met Einstein's photoelectric equation:

$hf = \phi + \tfrac{1}{2} m v_{\text{max}}^2$hf=ϕ+21​mvmax2​

This relates four physical quantities — photon frequency $f$f, work function $\phi$ϕ, electron mass $m$m and maximum photoelectron speed $v_{\text{max}}$vmax​ — together with Planck's constant $h$h. OCR H556 exam questions will ask you to find any one of these quantities, given the others, and also to compute the threshold frequency $f_0$f0​, the threshold wavelength $\lambda_0$λ0​ and the stopping potential $V_s$Vs​.

This lesson is a tour of the main calculation types you will encounter. Each example is worked out carefully, step by step, with particular attention to the unit conversions that cause so many careless errors. The closing pair of examples (calculations 5 and 9) constitute the PAG 5 protocol — the OCR-prescribed experimental method for determining Planck's constant from a stopping-potential graph. Practising these end-to-end is the single best preparation for both Paper 2 and Paper 3 photoelectric questions.

---

The Toolkit

List every relevant formula:

Quantity	Formula	Notes
Photon energy	$E_{\text{photon}} = hf = hc/\lambda$Ephoton​=hf=hc/λ	$E$E in J or eV
Einstein equation	$hf = \phi + KE_{\text{max}}$hf=ϕ+KEmax​	energy conservation
Max kinetic energy	$KE_{\text{max}} = hf - \phi$KEmax​=hf−ϕ	rearrangement
Max speed	$v_{\text{max}} = \sqrt{2 KE_{\text{max}}/m_e}$vmax​=2KEmax​/me​​	non-relativistic
Stopping potential	$eV_s = KE_{\text{max}}$eVs​=KEmax​	retarding p.d. zeroes the current
Threshold frequency	$f_0 = \phi/h$f0​=ϕ/h	$KE_{\text{max}} \to 0$KEmax​→0
Threshold wavelength	$\lambda_0 = c/f_0 = hc/\phi$λ0​=c/f0​=hc/ϕ	longest $\lambda$λ that emits
$V_s$Vs​-vs-$f$f graph	$V_s = (h/e) f - (\phi/e)$Vs​=(h/e)f−(ϕ/e)	slope $h/e$h/e, intercept $-\phi/e$−ϕ/e

Constants:

$h = 6.63 \times 10^{-34} \text{ J s} \qquad c = 3.00 \times 10^{8} \text{ m s}^{-1} \qquad e = 1.60 \times 10^{-19} \text{ C}$h=6.63×10−34 J sc=3.00×108 m s−1e=1.60×10−19 C

$m_e = 9.11 \times 10^{-31} \text{ kg} \qquad 1 \text{ eV} = 1.60 \times 10^{-19} \text{ J}$me​=9.11×10−31 kg1 eV=1.60×10−19 J

Shortcuts: $hc \approx 1.99 \times 10^{-25}$hc≈1.99×10−25 J m, equivalently $hc \approx 1240$hc≈1240 eV nm. The eV-nm form lets you compute photon energy in eV directly from wavelength in nm by simple division — by far the fastest route in stopping-potential problems.

---

Calculation 1: Finding the Work Function

Light of wavelength $350$350 nm falls on a metal surface; the maximum kinetic energy of the emitted photoelectrons is $1.70 \times 10^{-19}$1.70×10−19 J. Calculate the work function of the metal in both joules and electronvolts.

Solution. Photon energy:

$E_{\text{photon}} = \frac{hc}{\lambda} = \frac{1.99 \times 10^{-25}}{350 \times 10^{-9}} = 5.69 \times 10^{-19} \text{ J}$Ephoton​=λhc​=350×10−91.99×10−25​=5.69×10−19 J

Rearrange Einstein's equation:

$\phi = E_{\text{photon}} - KE_{\text{max}} = (5.69 \times 10^{-19}) - (1.70 \times 10^{-19}) = 3.99 \times 10^{-19} \text{ J}$ϕ=Ephoton​−KEmax​=(5.69×10−19)−(1.70×10−19)=3.99×10−19 J

Convert to eV:

$\phi = \frac{3.99 \times 10^{-19}}{1.60 \times 10^{-19}} \approx 2.49 \text{ eV}$ϕ=1.60×10−193.99×10−19​≈2.49 eV

Answer: $\phi \approx 4.0 \times 10^{-19}$ϕ≈4.0×10−19 J $\approx 2.5$≈2.5 eV (consistent with potassium or sodium).

---

Calculation 2: Threshold Frequency and Wavelength

A metal has work function $\phi = 4.50$ϕ=4.50 eV. Calculate (a) the threshold frequency and (b) the threshold wavelength.

Solution. Convert $\phi$ϕ to joules:

$\phi = (4.50)(1.60 \times 10^{-19}) = 7.20 \times 10^{-19} \text{ J}$ϕ=(4.50)(1.60×10−19)=7.20×10−19 J

(a) Threshold frequency:

$f_0 = \frac{\phi}{h} = \frac{7.20 \times 10^{-19}}{6.63 \times 10^{-34}} = 1.086 \times 10^{15} \text{ Hz}$f0​=hϕ​=6.63×10−347.20×10−19​=1.086×1015 Hz

(b) Threshold wavelength:

$\lambda_0 = \frac{c}{f_0} = \frac{3.00 \times 10^{8}}{1.086 \times 10^{15}} = 2.76 \times 10^{-7} \text{ m} = 276 \text{ nm}$λ0​=f0​c​=1.086×10153.00×108​=2.76×10−7 m=276 nm

Or equivalently, using the $1240$1240 eV nm shortcut directly:

$\lambda_0 = \frac{hc}{\phi} = \frac{1240}{4.50} \approx 276 \text{ nm}$λ0​=ϕhc​=4.501240​≈276 nm

Answer: $f_0 \approx 1.09 \times 10^{15}$f0​≈1.09×1015 Hz; $\lambda_0 \approx 276$λ0​≈276 nm.

This is in the far ultraviolet — visible light cannot liberate photoelectrons from this metal (consistent with zinc or aluminium, both around $\phi \sim 4.3$ϕ∼4.3 eV).

---

Calculation 3: Maximum Photoelectron Speed

UV light of wavelength $250$250 nm falls on a potassium surface ($\phi = 2.30$ϕ=2.30 eV). Calculate the maximum speed of the emitted photoelectrons.

Solution. Stay in eV until the kinetic-energy formula demands SI.

Photon energy: $E_{\text{photon}} = 1240/250 = 4.96$Ephoton​=1240/250=4.96 eV.

$KE_{\text{max}} = 4.96 - 2.30 = 2.66$KEmax​=4.96−2.30=2.66 eV.

Convert to joules:

$KE_{\text{max}} = (2.66)(1.60 \times 10^{-19}) = 4.26 \times 10^{-19} \text{ J}$KEmax​=(2.66)(1.60×10−19)=4.26×10−19 J

Apply $KE_{\text{max}} = (1/2) m_e v_{\text{max}}^2$KEmax​=(1/2)me​vmax2​:

$v_{\text{max}}^2 = \frac{2 KE_{\text{max}}}{m_e} = \frac{2 \times 4.26 \times 10^{-19}}{9.11 \times 10^{-31}} = 9.35 \times 10^{11}$vmax2​=me​2KEmax​​=9.11×10−312×4.26×10−19​=9.35×1011

$v_{\text{max}} = \sqrt{9.35 \times 10^{11}} \approx 9.67 \times 10^{5} \text{ m s}^{-1}$vmax​=9.35×1011​≈9.67×105 m s−1

Answer: $v_{\text{max}} \approx 9.7 \times 10^{5}$vmax​≈9.7×105 m s$^{-1}$−1, about $0.3\%$0.3% the speed of light. Non-relativistic, as expected.

---

Calculation 4: Stopping Potential

Monochromatic light of wavelength $310$310 nm illuminates a metal of work function $\phi = 3.00$ϕ=3.00 eV. Calculate the stopping potential.

Solution. Use $eV_s = KE_{\text{max}}$eVs​=KEmax​ and the convenient eV arithmetic.

Photon energy: $E_{\text{photon}} = 1240/310 \approx 4.00$Ephoton​=1240/310≈4.00 eV.

$KE_{\text{max}} = 4.00 - 3.00 = 1.00$KEmax​=4.00−3.00=1.00 eV.

Since $eV_s = KE_{\text{max}}$eVs​=KEmax​ and an electron crossing $1$1 V gains $1$1 eV of KE:

$V_s = 1.00 \text{ V}$Vs​=1.00 V

Answer: $V_s = 1.00$Vs​=1.00 V.

Exam Tip: When a question asks for the stopping potential and both the work function and photon energy are given (or easily computed) in eV, the answer is simply the difference in volts. No conversions to joules, no factors of $e$e: $V_s = (E_{\text{photon}}/\text{eV}) - (\phi/\text{eV})$Vs​=(Ephoton​/eV)−(ϕ/eV) gives $V_s$Vs​ directly in volts.

---

Calculation 5: Determining Planck's Constant from a Graph (PAG 5)

This is the PAG 5 anchor calculation — the OCR-prescribed experimental method for determining $h$h.

In a photoelectric experiment, the following data are collected:

$f$f / $10^{14}$1014 Hz	$V_s$Vs​ / V
5.5	0.16
6.5	0.58
7.5	0.99
8.5	1.40
9.5	1.82

Determine Planck's constant experimentally, and infer the work function of the metal.

Solution. Combining the Einstein equation with the stopping-potential relation:

$eV_s = hf - \phi \quad \Rightarrow \quad V_s = \left(\frac{h}{e}\right) f - \left(\frac{\phi}{e}\right)$eVs​=hf−ϕ⇒Vs​=(eh​)f−(eϕ​)

A plot of $V_s$Vs​ (vertical) against $f$f (horizontal) is therefore a straight line with slope $h/e$h/e and y-intercept $-\phi/e$−ϕ/e. The x-intercept is the threshold frequency $f_0 = \phi/h$f0​=ϕ/h.

Take two well-separated points: $(f_1, V_{s1}) = (5.5 \times 10^{14}, 0.16)$(f1​,Vs1​)=(5.5×1014,0.16) and $(f_2, V_{s2}) = (9.5 \times 10^{14}, 1.82)$(f2​,Vs2​)=(9.5×1014,1.82).

Gradient:

$\text{gradient} = \frac{V_{s2} - V_{s1}}{f_2 - f_1} = \frac{1.82 - 0.16}{(9.5 - 5.5) \times 10^{14}} = \frac{1.66}{4.0 \times 10^{14}} = 4.15 \times 10^{-15} \text{ V s}$gradient=f2​−f1​Vs2​−Vs1​​=(9.5−5.5)×10141.82−0.16​=4.0×10141.66​=4.15×10−15 V s

Planck's constant:

$h = (\text{gradient}) \times e = (4.15 \times 10^{-15})(1.60 \times 10^{-19}) = 6.64 \times 10^{-34} \text{ J s}$h=(gradient)×e=(4.15×10−15)(1.60×10−19)=6.64×10−34 J s

In excellent agreement with the accepted value $h = 6.63 \times 10^{-34}$h=6.63×10−34 J s — within $0.2\%$0.2%.

Work function: find the y-intercept by extrapolating the line back to $f = 0$f=0. Using point-slope from $(5.5 \times 10^{14}, 0.16)$(5.5×1014,0.16):

$V_s(\text{intercept}) = 0.16 - (4.15 \times 10^{-15})(5.5 \times 10^{14}) = 0.16 - 2.28 = -2.12 \text{ V}$Vs​(intercept)=0.16−(4.15×10−15)(5.5×1014)=0.16−2.28=−2.12 V

So $-\phi/e = -2.12$−ϕ/e=−2.12 V, giving $\phi = 2.12$ϕ=2.12 eV $\approx 3.4 \times 10^{-19}$≈3.4×10−19 J. This is consistent with caesium ($\phi \approx 2.1$ϕ≈2.1 eV).

Cross-check via threshold frequency: x-intercept at $V_s = 0$Vs​=0:

$0 = (4.15 \times 10^{-15}) f_0 - 2.12 \quad \Rightarrow \quad f_0 = \frac{2.12}{4.15 \times 10^{-15}} \approx 5.1 \times 10^{14} \text{ Hz}$0=(4.15×10−15)f0​−2.12⇒f0​=4.15×10−152.12​≈5.1×1014 Hz

And $\phi = h f_0 = (6.64 \times 10^{-34})(5.1 \times 10^{14}) \approx 3.4 \times 10^{-19}$ϕ=hf0​=(6.64×10−34)(5.1×1014)≈3.4×10−19 J $\approx 2.1$≈2.1 eV. ✓ Consistent.

Answer: $h \approx 6.6 \times 10^{-34}$h≈6.6×10−34 J s, $\phi \approx 2.1$ϕ≈2.1 eV. The graph yields both the universal constant ($h$h) and the metal-specific constant ($\phi$ϕ) from the same dataset — exactly the design intent of PAG 5.

This graphical data-handling question is a classic OCR format. It tests the Einstein equation, the stopping-potential relation, gradient and intercept analysis, and unit reasoning, all in one problem.

---

Calculation 6: Photocurrent and Intensity

A source emits light of wavelength $400$400 nm at a power of $0.10$0.10 mW. All of this light falls on a metal cathode of work function $2.00$2.00 eV, and every photon produces one photoelectron. Calculate (a) the number of photoelectrons emitted per second and (b) the resulting photocurrent in microamps.

Solution.

(a) Photon energy:

$E_{\text{photon}} = \frac{1240}{400} = 3.10 \text{ eV} = (3.10)(1.60 \times 10^{-19}) = 4.96 \times 10^{-19} \text{ J}$Ephoton​=4001240​=3.10 eV=(3.10)(1.60×10−19)=4.96×10−19 J

Number of photons per second (and, with unit quantum efficiency, photoelectrons per second):

$\frac{N}{t} = \frac{P}{E_{\text{photon}}} = \frac{1.0 \times 10^{-4}}{4.96 \times 10^{-19}} \approx 2.02 \times 10^{14} \text{ s}^{-1}$tN​=Ephoton​P​=4.96×10−191.0×10−4​≈2.02×1014 s−1

(b) Photocurrent:

$I = \frac{N}{t} \times e = (2.02 \times 10^{14})(1.60 \times 10^{-19}) \approx 3.23 \times 10^{-5} \text{ A} \approx 32 \text{ μA}$I=tN​×e=(2.02×1014)(1.60×10−19)≈3.23×10−5 A≈32 μA

Answer: $N/t \approx 2.0 \times 10^{14}$N/t≈2.0×1014 s$^{-1}$−1; $I \approx 32$I≈32 μA.

In practice, not every photon produces a photoelectron — the quantum efficiency of a real photocathode is typically only a few percent. The calculation above is the theoretical maximum, with each absorbed photon liberating exactly one photoelectron.

---

Calculation 7: Combining Photoelectric and Circuit Analysis

A clean metal surface has work function $\phi = 3.2$ϕ=3.2 eV. It is illuminated with ultraviolet light of wavelength $200$200 nm, and the photoelectrons are collected by an anode at potential $-2.0$−2.0 V relative to the cathode. Do any photoelectrons reach the anode? If so, what is their kinetic energy at the anode?

Solution. Photon energy: $E_{\text{photon}} = 1240/200 = 6.20$Ephoton​=1240/200=6.20 eV.

$KE_{\text{max}}$KEmax​ at the cathode: $6.20 - 3.20 = 3.00$6.20−3.20=3.00 eV.

The electrons must climb a potential hill of $2.0$2.0 V, costing them $eV = 2.0$eV=2.0 eV of kinetic energy. So at the anode they retain:

$KE_{\text{anode}} = 3.00 - 2.00 = 1.00 \text{ eV}$KEanode​=3.00−2.00=1.00 eV

Since this is positive, the fastest photoelectrons do reach the anode, arriving with $1.00$1.00 eV of kinetic energy.

If instead the retarding p.d. were $3.5$3.5 V, $KE_{\text{anode}} = 3.00 - 3.5 = -0.5$KEanode​=3.00−3.5=−0.5 eV — physically impossible — meaning no electrons reach the anode and the photocurrent is zero. The stopping potential is $V_s = 3.00$Vs​=3.00 V, the value of $KE_{\text{max}}$KEmax​ in electronvolts.

---

Calculation 8: Multi-wavelength Problem

A metal surface is sequentially illuminated with light of three wavelengths: $200$200 nm, $400$400 nm and $800$800 nm. The work function is $\phi = 3.0$ϕ=3.0 eV. In each case state whether emission occurs and (if so) calculate $KE_{\text{max}}$KEmax​.

Solution. Compute photon energy for each wavelength using $E(\text{eV}) = 1240/\lambda(\text{nm})$E(eV)=1240/λ(nm):

$\lambda$λ (nm)	$E_{\text{photon}}$Ephoton​ (eV)	Emission?	$KE_{\text{max}}$KEmax​ (eV)
200	6.20	Yes	$6.20 - 3.00 = 3.20$6.20−3.00=3.20
400	3.10	Yes	$3.10 - 3.00 = 0.10$3.10−3.00=0.10
800	1.55	No ($E_{\text{photon}} < \phi$Ephoton​<ϕ)	—

At $800$800 nm the photon energy ($1.55$1.55 eV) is less than the work function ($3.00$3.00 eV); no photoelectrons are emitted regardless of intensity. At $400$400 nm, emission just barely occurs — the fastest photoelectrons have only $0.10$0.10 eV ($\approx 1.6 \times 10^{-20}$≈1.6×10−20 J). At $200$200 nm, emission is robust.

flowchart LR
    L1[λ = 800 nm<br/>E = 1.55 eV] --> N1[No emission<br/>E < φ]
    L2[λ = 400 nm<br/>E = 3.10 eV] --> Y2[KE_max = 0.10 eV]
    L3[λ = 200 nm<br/>E = 6.20 eV] --> Y3[KE_max = 3.20 eV]

---

Calculation 9: PAG 5 Apparatus Errors

In a PAG 5 implementation, a student measures the stopping potential for monochromatic light from a discharge lamp using a filter. The student notes that the measured $V_s$Vs​ is systematically about $0.05$0.05 V smaller than expected. Suggest a probable cause and how it would affect the measured value of $h$h.

Solution. A common cause is stray light leakage at the filter — a small fraction of light at frequencies above the nominal value reaches the cathode, producing photoelectrons with higher $KE_{\text{max}}$KEmax​ than expected. Conversely, light below the nominal frequency (which should not produce emission) can leak through filter sidebands and contaminate the spectrum.

Another cause is thermionic emission: if the cathode is warm, electrons leave the metal even without photons, masking the true $V_s$Vs​. The graph would be shifted vertically and the gradient unchanged.

A third cause is contact potential: the cathode and anode are different metals (e.g. caesium vs tungsten) with different work functions; a contact potential of $\sim 0.5$∼0.5 V can appear in series with the applied p.d., giving a systematic offset. The gradient (and hence $h$h) is unchanged, but the y-intercept (and hence $\phi$ϕ) is shifted.

In all three cases the gradient of the $V_s$Vs​-vs-$f$f plot remains $h/e$h/e, so the measured Planck constant is unaffected by these systematic offsets. Only the intercept (and hence the inferred work function) shifts. This is why OCR mark schemes for PAG 5 typically require the candidate to quote $h$h from the gradient, not from any single point.

---

Method Summary: How to Approach a Photoelectric Question

flowchart TD
    A[Read the question] --> B{What is given?}
    B -- "λ" --> C[E_photon = hc/λ<br/>(use 1240/λ in eV nm shortcut)]
    B -- "f" --> D[E_photon = hf]
    B -- "V_s" --> E[KE_max = eV_s]
    C --> F{E_photon ≥ φ?}
    D --> F
    F -- "No" --> G[No emission, stop]
    F -- "Yes" --> H[KE_max = E_photon − φ]
    E --> H
    H --> I{Output asked?}
    I -- "V_s" --> J[V_s = KE_max/e<br/>(directly in V if KE in eV)]
    I -- "v_max" --> K[v_max = √(2 KE_max/m_e)<br/>(KE_max in J)]
    I -- "h or φ from graph" --> L[Plot V_s vs f<br/>Gradient × e = h<br/>y-intercept × e = −φ]