Photoelectric Effect Numericals for JEE Main

The photoelectric effect is one of the most reliable scoring chapters in JEE Main Modern Physics. It appears every year, and the numerical types are nearly identical across sessions. Once you have the five core formulas and their interrelationships in memory, every photoelectric numerical collapses into a substitution exercise. This guide works through each type systematically with the exact approach the exam rewards.

The Five Core Formulas

Every photoelectric numerical uses these five relations: (1) E = hν = hc/λ (photon energy); (2) KE_max = hν − φ (Einstein's equation, where φ is the work function); (3) eV₀ = KE_max (stopping potential definition); (4) ν₀ = φ/h (threshold frequency); (5) λ = h/p = h/√(2mKE) (de Broglie wavelength). Know the values h = 6.626 × 10⁻³⁴ J·s, hc = 1240 eV·nm (enormously useful), and e = 1.6 × 10⁻¹⁹ C. Memorise h/e = 4.14 × 10⁻¹⁵ eV·s to avoid unit conversions under pressure. For the broader modern physics context, see our modern physics guide.

Finding Threshold Wavelength and Work Function

If no photoelectrons are emitted for λ > 500 nm, then the threshold wavelength λ₀ = 500 nm and the work function φ = hc/λ₀ = 1240 eV·nm / 500 nm = 2.48 eV. This one-line calculation is the most common opening step. If the work function is given in electron-volts, convert to joules only for final substitution into the de Broglie formula (since KE then needs to be in joules for λ in metres). Working in eV throughout using hc = 1240 eV·nm eliminates most unit errors.

Maximum Kinetic Energy and Stopping Potential

For incident light of wavelength 300 nm on a surface with φ = 2 eV: photon energy = 1240/300 ≈ 4.13 eV; KE_max = 4.13 − 2 = 2.13 eV; stopping potential V₀ = 2.13 V. The stopping potential is numerically equal to KE_max when expressed in eV, so V₀ = KE_max/e in SI, or simply KE_max (in eV) = V₀ (in volts). This equivalence saves a conversion step. Notice that KE_max and V₀ are independent of intensity — only the number of photoelectrons depends on intensity.

Intensity Effects: A Conceptual Trap

Doubling the light intensity doubles the number of photoelectrons emitted per second (photocurrent) but does not change their maximum kinetic energy or the stopping potential. JEE frequently tests this as a conceptual MCQ. Intensity ∝ number of photons per second; frequency determines photon energy. Only increasing the frequency (and thus photon energy) increases KE_max. Confusing intensity with frequency effects is the trap in every conceptual photoelectric question.

De Broglie Wavelength from Stopping Potential

The de Broglie wavelength of the fastest photoelectrons follows from KE_max: λ_dB = h/√(2m·KE_max). Convert KE_max from eV to joules (multiply by 1.6 × 10⁻¹⁹), then substitute m = 9.1 × 10⁻³¹ kg. A useful shortcut: λ_dB (in nm) = 1.226/√(KE_max in eV). This shortcut works for electrons and is worth memorising for speed. After mastering these patterns, take a free mock test focused on modern physics to confirm your accuracy under time pressure.