Question

Light of wavelength 270 nm strikes a metal whose work function is 1.9 eV. What is the shortest de Broglie wavelength for the electrons that are produced as photoelectrons?

Answer 1

First find the energy "E" of the incoming light (photons) from;

E = hc/D , (D is the given wavelength in meters, h is Planck's constant in Joules-sec)

= (6.63*10^-34)(3*10^8)/(270*10^-9)

= 7.37*10^-19 J

The maximum kinetic energy of the emitted photoelectrons is the difference between the photon energy and the given work function;

KE = E - W = (7.37*10^-19) – (1.9*1.602*10^-19) = 4.32*10^-19 J

The maximum momentum of the photoelectrons is then;

P = SqRt(2m*KE), m is the electron mass.

= 8.87*10^-25 kg-m/s

Finally, the minimum DE Broglie wavelength "L" is;

L = h/P

= 7.47 x 10^-10 m