A light source of wavelength λ illuminates a metal and ejects photo-electrons with (K.E)max=1eV. Another light source of wavelength λ/3 ejects photo-electrons from the same metal with (K.E)max=4eV. Find the value of Work Function?
1 Answer
#phi = hnu_0 = "0.50 eV"#
As seen in the photoelectric effect, the kinetic energy
#K = hnu - hnu_0# where
#nu_0# is the threshold frequency, and#hnu_0# is otherwise known as the "work function".
From here we set up a system of equations to describe what is given in the question.
#"1 eV" = color(white)(.)(hc)/lambda - hnu_0# #" "" "" "bb((1))#
#"4 eV" = (3hc)/(lambda) - hnu_0# #" "" "" "bb((2))# where
#nu = c//lambda# for light waves and#lambda# is its wavelength.#c = 2.998 xx 10^8 "m/s"# is the speed of light.
Here we take
#" "-"3 eV" = -(3hc)/lambda + 3hnu_0#
#+ul(" ""4 eV" = " "(3hc)/(lambda) - color(white)(.)hnu_0)#
#" "color(white)(/)"1 eV" = " "" "" "" "2hnu_0#
So, the threshold energy is:
#color(blue)(phi = hnu_0 = "0.50 eV")#
As a check, does it work? From
#"1 eV" stackrel(?" ")(=) (hc)/lambda - "0.50 eV"#
#=> (hc)/lambda = "1.50 eV"#
Plugging this into
#"4 eV" stackrel(?" ")(=) (3hc)/lambda - "0.50 eV"#
#stackrel(?" ")(=) 3 ("1.50 eV") - "0.50 eV"#
#=# #"4 eV"# #color(blue)(sqrt"")#