Question

When ultraviolet 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 K0 of the photoelectrons when light of wavelength 310 nm falls on the same surface? Use h = 6.63×10−34 J⋅s for Planck's constant and c = 3.00×108 m/s for the speed of light and express your answer in electron volts.

Answer 1

You need to find the work function in order to solve this problem.
Also Planck's constant is also 4.14 * 10^-15 eV.

Max KE = 1.10 eV
lambda = 400 nm = 400 * 10^-9 m

First, find the frequency.

f = c / lambda
f = (3 * 10^8) / (400 * 10^-9 m)
f = 7.5 * 10^14

Max KE = hf - work function
1.10 eV = (4.14 * 10^-15 eV) (7.5 * 10^14) - work function
work function = 2.01 eV


Now find the frequency for the second wavelength.

f = c / lambda
f = (3 * 10^8) / (310 * 10^-9)
f = 9.68 * 10^14

Now plug this into your equation with this frequency and work function you found in the previous problem.

KEmax = hf - work function
KEmax = (4.14 * 10^-15 eV)(9.68 * 10^14) - 2.01 eV
KEmax = 1.9975 eV

Hope this helps