The 488 nm laser is shine on a lithium metal whose work function is 2.9 eV will you be able to see any photoelectrons? if yes what is the kinetic energy of the electrons if no what wavelength of light should be used to see the photoelectrons

There will not be any ejection of photoelectrons

Explanation:

Energy of the photon= hc/λ

Where;

h= Plank's constant

c= speed of light

λ= wavelength of the incident photon

E= 6.6×10^-34 × 3 ×10^8/488 × 10^-9

E= 4.1 ×10^-19 J

Work function of the metal (Wo)= 2.9 eV × 1.6 × 10^-19 = 4.64 × 10^-19 J

There can only be ejected photoelectrons when E>Wo but in this case, E<Wo hence there will not be any ejection of photoelectrons.