Ametal surface is illuminated by light of wavelength 310 nm. if the work function of the metal is 2.46 ev, what is the maximum kinetic energy of emitted photoelectrons? (for convenience, note that a photon of wavelength 1240 nm has an energy of 1 ev).a. 6.46 ev b. 4 ev c. 2.54 ev d. 1.46 ev e. 1.54 ev

(E) = 1.54ev

Explanation:

Maximum kinetic energy of emitted photoelectrons = hf - Ф

where h is Planck's constant = 6.63*10⁻³⁴ ,f is frequency = c/λ, c = velocity of light = 3.0*10⁸m/s, λ is wavelength of light = 310nm 310*10⁻⁹m, Ф = work function of metal = 2.46eV

max K.E. = hc/λ - Ф

max K.E. = (6.63*10⁻³⁴ * 3.0*10⁸m/s / 310nm 310*10⁻⁹m)  - 2.46eV

max K.E. = 6.416*10⁻¹⁹J - 2.46eV

Note; 1ev = 1.6*10⁻¹⁹J

max K.E. = {(6.416129*10⁻¹⁹J/ 1.6*10⁻¹⁹J) * 1eV} -2.46eV

max K.E. = 4.00eV -2.46eV

max K.E. = 1.54ev