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Photon-energy questions appear in both physics (dual nature, atoms) and chemistry (structure of atom). Doing them with 6.63 × 10⁻³⁴ and 3 × 10⁸ each time costs a minute and invites power-of-ten slips. The shortcut E = 1240/λ, with E in electronvolts and λ in nanometres, does the same job in one line. This post shows where 1240 comes from, how far you can trust it, the variants you will need, and the mistake it tempts you into.
Is it in the syllabus?
The physics is. The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus list the dual nature of radiation, the photoelectric effect and the particle nature of light in physics, and the nature of electromagnetic radiation and the hydrogen spectrum in chemistry. The NCERT chapters Dual Nature of Radiation and Matter (Class 12 physics) and Structure of Atom (Class 11 chemistry), both 2026-27 reprints, use E = hν = hc/λ with h = 6.63 × 10⁻³⁴ J s or 6.626 × 10⁻³⁴ J s. Neither prints the 1240 shortcut, so in a board answer, show E = hc/λ first.
Where does 1240 come from?
These three constants now have exact defined values. NIST lists h = 6.626 070 15 × 10⁻³⁴ J s, c = 299 792 458 m/s and e = 1.602 176 634 × 10⁻¹⁹ C. Then
hc/e = 1.239 84 × 10⁻⁶ V m = 1239.84 eV nm
So E(eV) = 1239.84/λ(nm), and 1240 is that number rounded. The rounding error is about 0.013%, far below the precision of any exam answer.
One thing to watch: NCERT's rounded constants (6.63 × 10⁻³⁴, 3 × 10⁸ and 1.6 × 10⁻¹⁹) give 1243 instead. If an answer key was built that way, your 1240 answer may differ in the third figure. Use the constants the question gives.
Worked examples
1. A green laser at 532 nm. E = 1240/532 ≈ 2.33 eV.
2. Checking a spectral line. NCERT's chemistry chapter gives the hydrogen electron energy as −2.18 × 10⁻¹⁸ J/n². The n = 3 to n = 2 jump releases 2.18 × 10⁻¹⁸ × (1/4 − 1/9) ≈ 3.03 × 10⁻¹⁹ J, or 1.89 eV. Then λ = 1240/1.89 ≈ 656 nm, the red Balmer line. Turning the formula around, λ = 1240/E, is often the faster direction.
3. X-rays. A wavelength of 0.1 nm gives 1240/0.1 = 12 400 eV, about 12.4 keV. With λ in ångströms, 1 Å = 0.1 nm, so E = 12 400/1 Å, the same answer.
4. Infrared. A CO₂ laser at 10.6 µm: λ = 10 600 nm, E ≈ 0.117 eV. Too little for most photoelectric surfaces.
5. Counting photons. A 5 mW pointer at 650 nm. Each photon has 1240/650 ≈ 1.91 eV = 1.91 × 1.602 × 10⁻¹⁹ ≈ 3.06 × 10⁻¹⁹ J. Photons per second = 5 × 10⁻³ ÷ 3.06 × 10⁻¹⁹ ≈ 1.6 × 10¹⁶. (Sample numbers for the pointer.)
6. Chemistry units. Multiply eV per photon by the Avogadro constant and e: 1 eV per particle is about 96.5 kJ/mol. So 532 nm light carries 2.33 × 96.5 ≈ 225 kJ per mole of photons, which you can compare with bond energies.
When does the shortcut give a wrong answer?
When the particle is not a photon. E = hc/λ uses the fact that a photon's energy is pc. For an electron, kinetic energy is p²/2m, so its de Broglie wavelength comes from λ = h/p, not from 1240/E. An electron with a de Broglie wavelength of 1 nm does not have 1240 eV of energy; it has about 1.5 eV. This is the most common misuse.
Which other mistakes cost marks?
λ in metres. 1240/(532 × 10⁻⁹) gives nonsense. Convert to nm first.
Leaving the answer in eV when joules are asked. Multiply by 1.602 × 10⁻¹⁹.
Using it for frequency. If you have ν, E = hν; or convert with λ = c/ν first.
Mixing NCERT's 1243-type constants with 1240 in one calculation.
For full photoelectric problems with work function and stopping potential, see our photoelectric effect problems post; for wavenumber and frequency conversions, our post on electromagnetic radiation calculations. If you are unsure whether a calculation is right after the shortcut, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning.
A practice routine
Day 1: Derive 1240 from h, c and e yourself, once with NIST values and once with NCERT's rounded ones.
Day 2: Ten conversions both ways (λ to E and E to λ), across radio, infrared, visible, ultraviolet and X-ray.
Day 3: Three photon-count questions and two kJ/mol questions.
Day 4: Two electron de Broglie questions, solved with λ = h/p, to train yourself out of the misuse.
Tonight, find the wavelength of a photon that carries exactly 3 eV and name its region of the spectrum.
