When ultraviolent light with a wavelength of 400 nm falls on

When ultraviolent light with a wavelength of 400 nm falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is 1.10 eV. What is the maximum kinetic energy K_0 of the photoelectrons when light of wavelength 270 nm falls on the same surface? Use h = 6.63 times 10^-34 J.s for Planck\'s constant and c = 3.00 times 10^8 m/s for the speed of light and express your answer in electron volts.

Solution

E = hV

= 6.62*10^-34*3*10^8)/(400*10^-9) = 4.97*10^-19 J = 3.1 ev

=> dE = 3.1 - 1.1 = 2 eV

at 270 nm

E2 = 6.62*10^-34*3*10^8)/(270*10^-9) = 7.3555556*10^-19 joule =  4.590 eV

Kelectron = 4.590 - 2 = 2.590 eV