A particular metal is illuminated by UV photons of wavelength 210 nm,and electrons up to 1.25 eV are emitted from the surface.What is the maximum KE of photoelectrons emitted when the UV photons are,instead,175 nm in wavelength?
A)1.18 eV
B)2.44 eV
C)5.90 eV
D)7.09 eV

Question

Quizplus · Accepted Answer

The energy of a photon is given by the equation E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon. 
For the original UV photons of wavelength 210 nm, the energy of the photon is: 
E = (6.626 x 10^-34 J s)(3.00 x 10^8 m/s)/(210 x 10^-9 m) = 2.98 eV
This means that the maximum KE of photoelectrons emitted is 1.73 eV (since energy is conserved and KE = E - work function of the metal). 
For the new UV photons of wavelength 175 nm: 
E = (6.626 x 10^-34 J s)(3.00 x 10^8 m/s)/(175 x 10^-9 m) = 3.57 eV 
The new maximum KE of photoelectrons emitted is 2.32 eV. 
Therefore, the correct answer is choice B, 2.44 eV, which is approximately equal to 1.73 eV + 0.71 eV (the difference between the energies of the two photons).

A Particular Metal Is Illuminated by UV Photons of Wavelength