Question

light of 380 nm wavelength is directed at a metal electrode. to determine the energy of electrons ejected, an opposing electrostatic potential difference is established between it and another electrode, the current of photoelectrons from one to the other is stopped completely when the potential difference is 1.1 V a) determine the work function of the metal b) determine the max wavelength light that can eject electron from this metal.

Answer 1

energy of electron ejected=1.1 eV

energy of the light with wavelength of 380 nm=h*c/wavelength

where h=plank's constant

c=speed of light

then energy=6.626*10^(-34)*3*10^8/(380*10^(-9))=5.2311*10^(-19) J

then in units of eV, energy=5.2311*10^(-19)/(1.6*10^(-19))=3.2694 eV

then work function=energy of the incident light wave-energy of the electron emitted

=3.2694-1.1=2.1694 eV

part b:

max wavelength light that can eject electron will be having minimum possible energy required to eject an electron i.e. work function of the metal.

==>if wavelength is lambda,

h*c/lambda=2.1694 eV

==>6.626*10^(-34)*3*10^8/lambda=2.1694*1.6*10^(-19)

==>lambda=6.626*10^(-34)*3*10^8/(2.1694*1.6*10^(-19))

==>lambda=5.7268*10^(-7) m=572.68 nm