Photoelectric Effect — Stopping potential, threshold frequency
Dual Nature of Matter
10
JEE Qs
8%
Hard
60
min
Master the graphs (K_max vs f, V_0 vs f, current vs intensity) and pay close attention to unit consistency (Joules vs eV) to avoid calculation errors.
🧮 Key Formulas
✅ Key Points for JEE
- 1The Photoelectric Effect confirms the particle nature of light (photons). Light energy is absorbed in discrete quanta (photons).
- 2The maximum kinetic energy (K_max) of emitted photoelectrons depends linearly on the frequency (f) of incident light, not its intensity.
- 3The photoelectric current is directly proportional to the intensity of incident light (for f > f_0) but is independent of its frequency.
- 4There exists a threshold frequency (f_0) for each material; no electron emission occurs if the incident frequency is below f_0, irrespective of light intensity. Correspondingly, there is a threshold wavelength (λ_0).
- 5Electron emission is instantaneous (no time lag) if the incident frequency is above the threshold frequency.
- 6The stopping potential (V_0) is the minimum negative potential applied to the collector plate to stop the most energetic photoelectrons; it is proportional to K_max and hence depends on the frequency of light and the material, but not on light intensity.
- 7Graphs of K_max vs f or V_0 vs f are straight lines. For K_max vs f, the slope is Planck's constant (h) and the y-intercept is -Φ (work function). For V_0 vs f, the slope is h/e and the x-intercept is f_0.
⚠️ Common Mistakes
- ✕Confusing the roles of intensity and frequency: Intensity affects the number of emitted electrons (current), while frequency affects the energy of individual electrons (K_max, V_0).
- ✕Incorrectly converting units (e.g., eV to Joules, Angstroms/nanometers to meters for wavelength) or using inconsistent units in calculations involving Planck's constant (h) or electron charge (e).
- ✕Ignoring the threshold condition (f < f_0 or λ > λ_0). If this condition is met, no photoemission occurs, and K_max (and V_0) should be zero, not negative.
📝 Practice Questions
See allQ42.A light source of wavelength λ illuminates a metal surface and electrons are ejected with maximum kinetic energy of 2 eV . If the same surface is illuminated by a light source of wavelength λ , then the maximum 2 kinetic energy of ejected electrons will be (The work function of metal is 1 eV ) (1) 3 eV (2) 2 eV (3) 6 eV (4) 5 eV
Q45.In photoelectric effect an em-wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is 2.14 eV and stopping potential is 2 V , what is the wavelength of the em- wave ? (Given hc = 1242eVnm where h is the Planck's constant and c is the speed of light in vaccum.) 2025 (23 Jan Shift 2) JEE Main Previous Year Paper (1) 300 nm (2) 400 nm (3) 600 nm (4) 200 nm
Q30.An electron in the ground state of the hydrogen atom has the orbital radius of 5.3 × 10−11 m while that for the electron in third excited state is 8.48 × 10−10 m . The ratio of the de Broglie wavelengths of electron in the excited state to that in the ground state is (1) 3 (2) 16 (3) 9 (4) 4
Q40.The work functions of cesium (Cs) and lithium (Li) metals are 1.9 eV and 2.5 eV , respectively. If we incident a light of wavelength 550 nm on these two metal surfaces, then photo-electric effect is possible for the case of (1) Both Cs and Li (2) Neither Cs nor Li (3) Cs only (4) Li only
Q34.A sub-atomic particle of mass 10−30 kg is moving with a velocity 2.21 × 106 m/s. Under the matter wave consideration, the particle will behave closely like (h = 6.63 × 10−34 J. s) (1) Visible radiation (2) Gamma rays (3) Infra-red radiation (4) X-rays
Q32.An electron of mass ' m ' with an initial velocity →v = v0^i (v0 > 0) enters an electric field E = −Eo^k . If the initial de Broglie wavelength is λ0 , the value after time t would be (1) λ0 (2) e2E20t2 m2v2o √1+ e2E02t2m2v02 λo√1 + (3) λ0 (4) λ0 √1−e2Eo2t2m2v2o
NCERT Chapters
- Class 12 Physics Ch 11: Dual Nature of Radiation and Matter