A surface whose work function is Phi = 4.4 eV is illuminated by a light whose wa

Question

A surface whose work function is Phi = 4.4 eV is illuminated by a light whose wavelength is 60.4 nm. What is the maximum kinetic energy of a photoelectron emitted from the surface? The speed of light is 3 times 10^8 m/s and Planck's constant is 6.63 times 10^-34 J middot s. Answer in units of eV. The light intensity incident on a metallic surface produces photoeletrons with a maximum kinetic energy of 3.3 eV. If the light intensity is doubled, what is the maximum kinetic energy of the photoeletrons? Answer in units of eV.

Explanation / Answer

here,

012)

work function , phi = 4.4 eV

wavelength of light , lamda = 60.4 * 10^-9 m

the maximum kinetic energy of photo electron emitted , E = initial energy - work function

E = h * c/( lamda) - phi * e

E = 6.626 * 10^-34 * 3 * 10^8/( 60.4 * 10^-9) - ( 4.4 * 1.6 * 10^-19)

E = 2.58 * 10^-18 J

E = 16.1 eV

the maximum kinetic energy of the photoelectrons is 16.1 eV

013)

when the intensity of the light is doubled,

the maximum kinetic energy od the photoelectrons remains the same,

only the number of electrons changes

so, the kinetic energy is 3.3 eV

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