Answer:a) 492 nm; b) 1.78 eV
Explanation: In order to solve this problem we have to use the photoelectric energy balance givenb by:
h*ν=Ek+W where h is the Placnk constant, ν is the frequency of the radiation, Ek is the kinetic energy of the realised electrons and W is the work funcion of the incident metal.
Then we have:
W=h*ν-Ek= h*c/λ-Ek=1240 eV.nm/176 nm=2.52 eV
so the maximun wavelength to realese the electrons is when Ek=0 then
W=h*c/λmax so λmax= h*c/W= 1240/2.52=492.06 nm
Finally if we use 288 nm to realease the electrons, then Ek of the emitted electron from the metal is:
from h*ν=Ek+W we have:
Ek=h*c/λ-W= (1240/288) eV*nm-2.52 eV=1.78 eV