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"How many photons per second does this laser emit?" is a one-line question that hides three unit traps and one physics trap. It also leads straight into harder questions: how many photons reach a detector, or what current they produce. This post builds the method, works four cases with full numbers, and lists the slips to avoid.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus both list the photoelectric effect and the "particle nature of light" under dual nature of matter and radiation. The textbook is section 11.7, Particle Nature of Light: The Photon, in NCERT Class 12 Physics, Dual Nature of Radiation and Matter (2026-27 reprint). Its Example 11.1 is a laser photon count. Check your exam year's syllabus.
What is the one equation?
NCERT's photon picture says each photon of frequency ν has energy E = hν, all photons of one wavelength have the same energy, and raising the intensity only raises the number of photons per second. So if a source emits light power P:
P = N × E, so N = P/E = Pλ/(hc)
The constants are exact in SI. 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. From these, hc = 1239.8 eV nm, so a handy shortcut is E (in eV) ≈ 1240 ÷ λ (in nm).
Worked example 1: a green laser pointer
Sample numbers: 5.0 mW of light at 532 nm.
E = hc/λ = (6.626 × 10⁻³⁴ × 2.998 × 10⁸) ÷ (532 × 10⁻⁹) = 3.73 × 10⁻¹⁹ J. Check: 1240 ÷ 532 = 2.33 eV.
N = P/E = 5.0 × 10⁻³ ÷ 3.73 × 10⁻¹⁹ = 1.34 × 10¹⁶ photons per second.
Worked example 2: same power, red light
Now 5.0 mW at 650 nm. Each photon has 1240 ÷ 650 = 1.91 eV, or 3.06 × 10⁻¹⁹ J, so N = 1.64 × 10¹⁶ per second.
The red beam has 650 ÷ 532 ≈ 1.22 times as many photons. At equal power, longer wavelength means less energy per photon and so more photons. Students who think "red is weaker, so fewer photons" get this backwards.
Worked example 3: a lamp, and how many reach your eye
Sample numbers: a sodium lamp emits 12 W of light at 589 nm, equally in all directions. You stand 10 m away, and your pupil has a radius of 2.0 mm.
E = 1240 ÷ 589 = 2.11 eV, or 3.37 × 10⁻¹⁹ J.
N = 12 ÷ 3.37 × 10⁻¹⁹ = 3.56 × 10¹⁹ photons per second from the lamp.
At 10 m the light is spread over a sphere of area 4π × 10² = 1257 m².
Pupil area: π × (2.0 × 10⁻³)² = 1.26 × 10⁻⁵ m².
Fraction entering the eye: 1.26 × 10⁻⁵ ÷ 1257 = 1.0 × 10⁻⁸.
Photons into the eye: 3.56 × 10¹⁹ × 1.0 × 10⁻⁸ ≈ 3.6 × 10¹¹ per second.
This ignores absorption in air. The 12 W is light power: a lamp rated by electrical input emits less as light, so a question must tell you the light output or an efficiency.
Worked example 4: from photons to photocurrent
Sample numbers: 2.0 μW of 400 nm light falls on a photocathode, and 1 photon in 10 ejects an electron. All ejected electrons are collected.
E = 1240 ÷ 400 = 3.10 eV, or 4.97 × 10⁻¹⁹ J.
Photons per second: 2.0 × 10⁻⁶ ÷ 4.97 × 10⁻¹⁹ = 4.03 × 10¹².
Electrons per second: 4.03 × 10¹¹.
Current: 4.03 × 10¹¹ × 1.602 × 10⁻¹⁹ = 6.5 × 10⁻⁸ A, about 65 nA.
This matches NCERT's experimental result that photocurrent is directly proportional to the intensity of light above the threshold: double the photons, double the current. The wavelength must also be short enough for the metal, as in our post on photoelectric effect problems.
Which mistakes cost marks?
Leaving λ in nanometres inside hc/λ when h and c are in SI units. Convert to metres, or use the 1240 eV nm shortcut and convert eV to joules.
Dividing power in watts by energy in eV. Units must match: 1 eV = 1.602 × 10⁻¹⁹ J.
Using electrical power instead of emitted light power.
Forgetting the geometry. Only a fraction of an isotropic source's photons reach a small area.
Rounding hc too early. With 1240 the error is about 0.01 per cent, which is fine; with 1200 it is over 3 per cent.
A practice routine
Solve NCERT exercise 11.4: a 9.42 mW helium-neon laser at 632.8 nm. You should find about 3.0 × 10¹⁶ photons per second.
Rework example 3 at 20 m and confirm the count falls by a factor of 4.
Rework example 4 for 600 nm light at the same power and efficiency, then explain the change in current.
Write the shortcut E ≈ 1240/λ from memory and derive it from NIST's h, c and e.
If a step does not make sense, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer. For the force light exerts, see our post on photon momentum and radiation pressure.
Tonight, estimate the photons per second from your phone's torch, assuming 0.1 W of light at 550 nm.
