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:

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