When ultraviolet light with a wavelength of 400 nm falls on a certain metal surf

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 K_0 of the photoelectrons when light of wavelength 330 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.

Other solved answers are out there but I have tried them and they come up wrong for this problem. A completely worked out problem would be amazingly helpful.

Explanation / Answer

Wavelength of an ultraviolet light is = 400*10^-9 m Maximum kinetic energy is K.E = 1.10eV KE _max = hf -      = hf - KE_max          = hc - K.E max /          = 6.625 *10^-34 (3.00*10^8 m/s )- 1.10*1.6*10^-19 J / 400*10^-9 m          = 3.20*10^-19 J Kinetic energy for the lowest wavelength is    K.E = h c - /            = 6.625 *10^-34 (3.00*10^8 m/s)   - 3.20*10^-19 J / 330*10^-9 m           = 2.82*10^-19 J           = 1.76 eV

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