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
Truong-Son N.
May 31, 2018

#phi = hnu_0 = "0.50 eV"#

As seen in the photoelectric effect, the kinetic energy #K# is leftover from the incoming light source energy #hnu# that is in excess of the threshold energy #hnu_0#:

#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 #-3cdot(1) + (2)# to get:

#" "-"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)#:

#"1 eV" stackrel(?" ")(=) (hc)/lambda - "0.50 eV"#

#=> (hc)/lambda = "1.50 eV"#

Plugging this into #(2)#:

#"4 eV" stackrel(?" ")(=) (3hc)/lambda - "0.50 eV"#

#stackrel(?" ")(=) 3 ("1.50 eV") - "0.50 eV"#

#=# #"4 eV"# #color(blue)(sqrt"")#

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