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 Ko of the photoelectrons when light of wavelength 320 nm falls on the same surface?

Use h = 6.63

Answer 1

Wavelength ? = 400 nm = 400*10^-9 m = 4*10^-7 m

Energy of photon E = hc/? = (6.63*10^-34*3*10^8)/(4*10^-7)

                                       = 4.972*10^-19 J = (4.972*10^-19)/(1.602*10^-19)

                                                                = 3.103 eV

Maximum kinetic energy Ko = 1.10 eV

Work function of the metal surface W = E - Ko = 3.103 - 1.10 = 2.003 eV

New wavelength ?' = 320 nm = 320*10^-9 m

Energy of the incident photon E' = 6.416*10^-19 J = 4.004 eV

Maximium Kinetic energy Ko' = E' - W

                                          = 4.004 - 2.003 = 2.001 eV = 3.205*10^-19 J