The work function of caesium metal is 2.14 eV. When light of frequency 10^{14}$$ Hz is incident on the metal surface, photoemission of electrons occurs. What is the (a) maximum kinetic energy of the emitted electrons, (b) Stopping potential, and (c) maximum speed of the emitted photoelectrons?
Maximum kinetic energy of emitted electrons = 0.34 eV
Stopping potential = 0.34 V
Maximum speed of emitted photoelectrons = ⁵ m/s
Explanation
This question applies Einstein's photoelectric equation K_max = hν - φ₀. The work function φ₀ = 2.14 eV represents the minimum energy needed for electron emission. When light of frequency ν = ¹⁴ Hz strikes the caesium surface, photons transfer energy to electrons. The excess energy beyond the work function appears as kinetic energy of emitted electrons. The stopping potential equals the maximum kinetic energy in electron-volts, and the speed is calculated using the classical kinetic energy formula.