Light of wavelength 400 nm is incident continuously on a cesium ball (work function 1.9 eV). The maximum potential to which the ball will be charged is:
Correct answer: B. 1.2 V
- A. 3.1 V
- B. 1.2 V
- C. Zero
- D. Infinite
Explanation
Calculate the energy of the incident photon:Use the formula: Ephoton = (hc) / λWhere:Ephoton is the energy of the photonh is Planck's constant (6.626 x 10-34 J s)c is the speed of light (3 x 108 m/s)λ is the wavelength of the light (400 nm = 400 x 10-9 m)E_photon = (6.626 x 10-34 J s * 3 x 108 m/s) / (400 x 10-9 m)E_photon = 4.9695 x 10-19 J Convert the photon energy to electron volts (eV):1 eV = 1.602 x 10-19JE_photon (eV) = E_photon (J) / (1.602 x 10-19 J/eV)E_photon (eV) = 4.9695 x 10-19 J / (1.602 x 10-19J/eV)E_photon (eV) ≈ 3.1 eVUse the photoelectric effect equation:E_photon = Work Function (Φ) + Kinetic Energy (KE) of emitted electronsCalculate the maximum kinetic energy of the emitted electrons:KEmax = E_photon - ΦKEmax = 3.1 eV - 1.9 eVKEmax = 1.2 eVDetermine the stopping potential (V_s):The maximum kinetic energy of the emitted electrons is equal to the work done by the stopping potential to stop them.KEmax = eV_s, where e is elementary charge (i.e. we divide KE by electron charge) and Vs is the stopping potentialVs = KEmax / e, and since the KE is already in eV, then it will simply be:Vs = 1.2 V
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Classical physics cannot explain blackbody radiation and related observations, leading to Planck’s quantum theory, in which energy is emitted or absorbed in discrete packets. Photons provide the particle model of light, with energy and momentum linked to frequency and wavelength, including the photoelectric effect and its threshold frequency.
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