The photoelectric work - function of potassium is 2.3 eV. If light
having a wavelength of `2800 Å` falls on potassium, find
(a) the kinetic energy in electron volts of the most energetic electrons ejected.
(b) the stopping potential in volts.

Explanation

Given, `W=2.3eV" "lamda=2800Å`
`:.E("in "eV)=(12400)/(lamda("in"Å))=(12400)/(2800)=4.4eV`
`K_(max)=E-W=(4.4-2.3)eV=2.1eV`
(b) `K_(max)=eV_(0):.2.1eV=eV_(0)(or)V_(0)=2.1` volt