Photon Energy Calculations

Spec mapping: OCR H556 Module 4.5 — Quantum physics (numerical use of $E = hf = hc/\lambda$E=hf=hc/λ; conversion between joules and electronvolts; photon-flux calculations $N/t = P/E_{\text{photon}}$N/t=P/Ephoton​). Refer to the official OCR H556 specification document for exact wording.

In the previous lesson we introduced the photon model and the Planck–Einstein relation $E = hf = hc/\lambda$E=hf=hc/λ. You should now be comfortable with the idea that electromagnetic energy is quantised. This lesson is about doing quantum arithmetic with confidence: taking a wavelength or a frequency or an energy and computing the other two, converting between joules and electronvolts, and comparing photons across the electromagnetic spectrum. OCR H556 exam questions rarely use the Planck relation in only one direction, and they frequently combine it with other ideas from the specification — conservation of energy, $c = f\lambda$c=fλ, the definition of the electronvolt, and (foreshadowing lessons 3–5) the photoelectric threshold.

The material in this lesson is methodological rather than conceptual. It builds a toolkit of nine worked examples and a clean unit-discipline workflow you can draw on whenever the exam asks for a numerical answer in eV, J, keV, MeV or photons per second.

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Recap: The Core Equations

Before calculating, collect every equation in one place.

Quantity	Equation	Units / notes
Photon energy (from frequency)	$E = hf$E=hf	$E$E in J, $f$f in Hz
Photon energy (from wavelength)	$E = hc/\lambda$E=hc/λ	$E$E in J, $\lambda$λ in m
Wave equation in vacuum	$c = f\lambda$c=fλ	$c = 3.00 \times 10^{8}$c=3.00×108 m s$^{-1}$−1
Energy unit conversion	$1 \text{ eV} = 1.60 \times 10^{-19}$1 eV=1.60×10−19 J	divide / multiply by $e$e
Photon flux from total power	$N/t = P/E_{\text{photon}}$N/t=P/Ephoton​	photons per second
Threshold wavelength (lesson 4)	$\lambda_0 = hc/\phi$λ0​=hc/ϕ	minimum to liberate electron

The OCR data-sheet constants used throughout:

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

Two derived shortcuts save enormous time once committed to memory:

$hc = 1.99 \times 10^{-25} \text{ J m} \qquad hc \approx 1240 \text{ eV nm}$hc=1.99×10−25 J mhc≈1240 eV nm

The first form pairs with wavelength in metres and answer in joules; the second pairs with wavelength in nanometres and answer in electronvolts. Choose the form that minimises unit gymnastics — this is the secret to fast, error-free photon arithmetic.

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Worked Example 1: Visible Blue Light

A photon of blue light has wavelength $\lambda = 450$λ=450 nm. Calculate its energy in (a) joules and (b) electronvolts.

Solution.

(a) Using $E = hc/\lambda$E=hc/λ with $hc = 1.99 \times 10^{-25}$hc=1.99×10−25 J m and $\lambda = 450 \times 10^{-9}$λ=450×10−9 m:

$E = \frac{1.99 \times 10^{-25}}{450 \times 10^{-9}} = 4.42 \times 10^{-19} \text{ J}$E=450×10−91.99×10−25​=4.42×10−19 J

(b) Convert to eV by dividing by $e$e:

$E = \frac{4.42 \times 10^{-19}}{1.60 \times 10^{-19}} = 2.76 \text{ eV}$E=1.60×10−194.42×10−19​=2.76 eV

Equivalently using $E(\text{eV}) = 1240/\lambda(\text{nm}) = 1240/450 = 2.76$E(eV)=1240/λ(nm)=1240/450=2.76 eV — same answer in one line.

Answer: $E \approx 4.4 \times 10^{-19}$E≈4.4×10−19 J $\approx 2.8$≈2.8 eV.

Exam Tip: Notice how quickly the calculation proceeds once you have memorised $hc = 1.99 \times 10^{-25}$hc=1.99×10−25 J m (or $1240$1240 eV nm). Recomputing $(6.63 \times 10^{-34})(3.00 \times 10^{8})$(6.63×10−34)(3.00×108) every time is unnecessary work and risks arithmetic slips.

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Worked Example 2: Infrared Radiation

An infrared photon has frequency $f = 3.75 \times 10^{13}$f=3.75×1013 Hz. Calculate (a) its energy in joules, (b) in electronvolts, and (c) its wavelength.

Solution.

(a) $E = hf = (6.63 \times 10^{-34})(3.75 \times 10^{13}) = 2.49 \times 10^{-20}$E=hf=(6.63×10−34)(3.75×1013)=2.49×10−20 J.

(b) $E = (2.49 \times 10^{-20})/(1.60 \times 10^{-19}) = 0.155$E=(2.49×10−20)/(1.60×10−19)=0.155 eV.

(c) $\lambda = c/f = (3.00 \times 10^{8})/(3.75 \times 10^{13}) = 8.00 \times 10^{-6}$λ=c/f=(3.00×108)/(3.75×1013)=8.00×10−6 m $= 8.00$=8.00 μm.

This is mid-infrared — the characteristic wavelength of thermal radiation from objects near room temperature, exactly the band detected by a thermal imaging camera.

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Worked Example 3: Comparing Red and Violet Light

Red light has wavelength $700$700 nm. Violet light has wavelength $400$400 nm. Calculate the ratio $E_{\text{violet}} : E_{\text{red}}$Eviolet​:Ered​.

Solution. Because $E = hc/\lambda$E=hc/λ, photon energy is inversely proportional to wavelength:

$\frac{E_{\text{violet}}}{E_{\text{red}}} = \frac{\lambda_{\text{red}}}{\lambda_{\text{violet}}} = \frac{700}{400} = 1.75$Ered​Eviolet​​=λviolet​λred​​=400700​=1.75

A violet photon carries $1.75$1.75 times the energy of a red photon. Numerical check:

$E_{\text{red}} = \frac{1.99 \times 10^{-25}}{700 \times 10^{-9}} = 2.84 \times 10^{-19} \text{ J} = 1.78 \text{ eV}$Ered​=700×10−91.99×10−25​=2.84×10−19 J=1.78 eV

$E_{\text{violet}} = \frac{1.99 \times 10^{-25}}{400 \times 10^{-9}} = 4.98 \times 10^{-19} \text{ J} = 3.11 \text{ eV}$Eviolet​=400×10−91.99×10−25​=4.98×10−19 J=3.11 eV

and indeed $3.11/1.78 \approx 1.75$3.11/1.78≈1.75.

Exam Tip: OCR questions frequently ask for ratios of photon energies across different wavelengths. The ratio comes out directly from $E \propto 1/\lambda$E∝1/λ without needing either absolute value separately. This is often the quickest route.

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Worked Example 4: Counting Photons in a Laser Beam

A $5.0$5.0 mW laser produces green light of wavelength $532$532 nm. How many photons does the laser emit per second?

Solution.

Step 1. Energy of one photon:

$E_{\text{photon}} = \frac{hc}{\lambda} = \frac{1.99 \times 10^{-25}}{532 \times 10^{-9}} = 3.74 \times 10^{-19} \text{ J}$Ephoton​=λhc​=532×10−91.99×10−25​=3.74×10−19 J

Step 2. Power $P = 5.0$P=5.0 mW delivers $5.0 \times 10^{-3}$5.0×10−3 J per second. Photon flux:

$\frac{N}{t} = \frac{P}{E_{\text{photon}}} = \frac{5.0 \times 10^{-3}}{3.74 \times 10^{-19}} = 1.34 \times 10^{16} \text{ s}^{-1}$tN​=Ephoton​P​=3.74×10−195.0×10−3​=1.34×1016 s−1

Answer: about $1.3 \times 10^{16}$1.3×1016 photons per second.

This is an enormous number — which is why the beam looks continuous. The eye can no more count individual photons than count raindrops in a downpour.

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Worked Example 5: X-Ray Photon Energy

A medical X-ray source produces photons of wavelength $\lambda = 5.0 \times 10^{-11}$λ=5.0×10−11 m (i.e. $0.05$0.05 nm). Calculate the photon energy in keV.

Solution.

$E = \frac{hc}{\lambda} = \frac{1.99 \times 10^{-25}}{5.0 \times 10^{-11}} = 3.98 \times 10^{-15} \text{ J}$E=λhc​=5.0×10−111.99×10−25​=3.98×10−15 J

Convert to electronvolts:

$E = \frac{3.98 \times 10^{-15}}{1.60 \times 10^{-19}} = 2.49 \times 10^{4} \text{ eV} = 24.9 \text{ keV}$E=1.60×10−193.98×10−15​=2.49×104 eV=24.9 keV

A $25$25 keV photon is characteristic of a diagnostic X-ray machine — enough to penetrate soft tissue, ionise atoms and produce a radiographic image, but not so energetic as to damage tissue uniformly throughout the body.

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Worked Example 6: Reverse Problem — Wavelength from Energy

A photon has energy $3.40$3.40 eV. What is its wavelength?

Solution. Quickest route: the $1240$1240 eV nm shortcut.

$\lambda(\text{nm}) = \frac{1240}{E(\text{eV})} = \frac{1240}{3.40} = 365 \text{ nm}$λ(nm)=E(eV)1240​=3.401240​=365 nm

Cross-check in SI units: $E = (3.40)(1.60 \times 10^{-19}) = 5.44 \times 10^{-19}$E=(3.40)(1.60×10−19)=5.44×10−19 J, so

$\lambda = \frac{hc}{E} = \frac{1.99 \times 10^{-25}}{5.44 \times 10^{-19}} = 3.66 \times 10^{-7} \text{ m} = 366 \text{ nm}$λ=Ehc​=5.44×10−191.99×10−25​=3.66×10−7 m=366 nm

The $1$1 nm difference is rounding noise. The answer lies in the near-ultraviolet, just beyond the violet edge of the visible spectrum.

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Joules versus Electronvolts: A Conversion Table

From	To	Operation	Worked example
J	eV	divide by $1.60 \times 10^{-19}$1.60×10−19	$3.2 \times 10^{-19}$3.2×10−19 J $\to 2.0$→2.0 eV
eV	J	multiply by $1.60 \times 10^{-19}$1.60×10−19	$2.0$2.0 eV $\to 3.2 \times 10^{-19}$→3.2×10−19 J
J	keV	divide by $1.60 \times 10^{-16}$1.60×10−16	$3.98 \times 10^{-15}$3.98×10−15 J $\to 24.9$→24.9 keV
J	MeV	divide by $1.60 \times 10^{-13}$1.60×10−13	$2.0 \times 10^{-13}$2.0×10−13 J $\to 1.25$→1.25 MeV
eV	keV	divide by $10^{3}$103	$24{,}900$24,900 eV $\to 24.9$→24.9 keV
eV	MeV	divide by $10^{6}$106	$1{,}250{,}000$1,250,000 eV $\to 1.25$→1.25 MeV

When the exam quotes an answer in electronvolts but your formula uses joules, the safest workflow is: calculate in joules throughout, then convert at the end. Mixing units partway through is a reliable recipe for errors. Alternatively, use $hc \approx 1240$hc≈1240 eV nm from the start and stay in eV throughout.

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Comparing Photons Across the Spectrum

flowchart LR
    A["Radio<br/>λ = 1 m<br/>E = 1.2 × 10⁻²⁵ J<br/>= 7.5 × 10⁻⁷ eV"]
    B["Microwave<br/>λ = 10 mm<br/>E = 2 × 10⁻²³ J<br/>= 1.2 × 10⁻⁴ eV"]
    C["Visible<br/>λ = 550 nm<br/>E = 3.6 × 10⁻¹⁹ J<br/>= 2.3 eV"]
    D["UV<br/>λ = 100 nm<br/>E = 2.0 × 10⁻¹⁸ J<br/>= 12 eV"]
    E["X-ray<br/>λ = 0.1 nm<br/>E = 2.0 × 10⁻¹⁵ J<br/>= 12 keV"]
    F["Gamma<br/>λ = 0.001 nm<br/>E = 2.0 × 10⁻¹³ J<br/>= 1.2 MeV"]
    A --> B --> C --> D --> E --> F

A single radio photon carries less than a millionth of an electronvolt; a single gamma photon carries more than a million electronvolts. Twelve orders of magnitude separate the two ends of the electromagnetic spectrum, and those twelve orders of magnitude are precisely why radio waves bounce harmlessly off our skin while gamma rays ionise our DNA.

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Worked Example 7: Photon Flux from a 60 W Bulb

A $60$60 W incandescent light bulb emits about $10\%$10% of its electrical power as visible light (the rest is infrared). Estimate how many visible photons it emits per second, treating all visible photons as having the energy of $550$550 nm green light.

Solution.

Visible power: $P_{\text{vis}} = 0.10 \times 60 = 6.0$Pvis​=0.10×60=6.0 W.

Photon energy: $E_{\text{ph}} = hc/\lambda = (1.99 \times 10^{-25})/(550 \times 10^{-9}) = 3.62 \times 10^{-19}$Eph​=hc/λ=(1.99×10−25)/(550×10−9)=3.62×10−19 J.

Photons per second:

$\frac{N}{t} = \frac{P_{\text{vis}}}{E_{\text{ph}}} = \frac{6.0}{3.62 \times 10^{-19}} \approx 1.66 \times 10^{19} \text{ s}^{-1}$tN​=Eph​Pvis​​=3.62×10−196.0​≈1.66×1019 s−1

Almost $2 \times 10^{19}$2×1019 visible photons every second. The eye averages over their discrete arrival times completely; the lamp appears to glow with smooth, continuous light.

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Worked Example 8: Minimum Wavelength to Ionise Hydrogen

The ionisation energy of hydrogen is $13.6$13.6 eV. What is the longest wavelength of light (lowest photon energy) that can ionise a hydrogen atom?

Solution. Minimum photon energy is $13.6$13.6 eV; maximum wavelength corresponds. Using $E(\text{eV}) = 1240/\lambda(\text{nm})$E(eV)=1240/λ(nm) in reverse:

$\lambda_{\text{max}} = \frac{1240}{13.6} \approx 91 \text{ nm}$λmax​=13.61240​≈91 nm

Cross-check in SI: $E = (13.6)(1.60 \times 10^{-19}) = 2.18 \times 10^{-18}$E=(13.6)(1.60×10−19)=2.18×10−18 J; $\lambda_{\text{max}} = hc/E = (1.99 \times 10^{-25})/(2.18 \times 10^{-18}) = 9.13 \times 10^{-8}$λmax​=hc/E=(1.99×10−25)/(2.18×10−18)=9.13×10−8 m $\approx 91$≈91 nm.

This is in the extreme ultraviolet (the Lyman series limit). Photons of longer wavelength — visible, infrared, radio — cannot ionise hydrogen no matter how many of them you throw at it. This is exactly the kind of threshold behaviour we meet in the photoelectric effect in lessons 3 and 4.

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Worked Example 9: Multi-Wavelength Photon Inventory

A street lamp emits, in equal power at each wavelength, photons at $589$589 nm (sodium yellow), $436$436 nm (mercury blue) and $405$405 nm (mercury violet). If the total radiant power is $30$30 W (visible only) split equally between the three wavelengths, calculate the photon flux at each wavelength.

Solution. Power per wavelength: $P_i = 30/3 = 10$Pi​=30/3=10 W.

$\lambda$λ (nm)	$E_{\text{ph}}$Eph​ (eV)	$E_{\text{ph}}$Eph​ (J)	$N_i/t = P_i/E_{\text{ph}}$Ni​/t=Pi​/Eph​ (s$^{-1}$−1)
589	$1240/589 = 2.10$1240/589=2.10	$3.37 \times 10^{-19}$3.37×10−19	$2.97 \times 10^{19}$2.97×1019
436	$1240/436 = 2.84$1240/436=2.84	$4.55 \times 10^{-19}$4.55×10−19	$2.20 \times 10^{19}$2.20×1019
405	$1240/405 = 3.06$1240/405=3.06	$4.90 \times 10^{-19}$4.90×10−19	$2.04 \times 10^{19}$2.04×1019

Each wavelength gives a slightly different photon flux because each photon carries a different energy. The shorter the wavelength, the more energetic each photon and (for equal power) the fewer photons per second. Total photon flux: about $7.2 \times 10^{19}$7.2×1019 s$^{-1}$−1.

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Method Summary: A Photon-Calculation Checklist

flowchart TD
    A[Read the question] --> B{What is given?}
    B -- "λ or f" --> C[Compute E_photon<br/>via hc/λ or hf]
    B -- "E (J or eV)" --> D[Compute f or λ<br/>via E = hf or hc/λ]
    B -- "Power P" --> E[Compute photon flux<br/>N/t = P/E_photon]
    C --> F{Output unit asked?}
    D --> F
    E --> F
    F -- "J" --> G[Stay in SI]
    F -- "eV" --> H[Divide by 1.60 × 10⁻¹⁹<br/>OR use 1240/λ shortcut]
    F -- "keV / MeV" --> I[Scale by 10³ / 10⁶]
    G --> J[Sanity-check: visible ≈ 10⁻¹⁹ J;<br/>X-ray ≈ 10⁻¹⁵ J;<br/>gamma ≈ 10⁻¹³ J]
    H --> J
    I --> J

If your answer's order of magnitude is out by $10^9$109, suspect a missing nm $\to$→ m conversion. If it's out by $10^{19}$1019, suspect a missing J $\leftrightarrow$↔ eV conversion.

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