What is the wavelength of light that hits potassium metal in order to eject photoelectrons with a wavelength of #"0.57 nm"#? The ionization energy of potassium is #"4.34 eV"#.

1 Answer
Sep 16, 2017

I got ultraviolet light of about #"138.24 nm"#. If you got #"0.57 nm"#, you assumed the photoelectrons had no energy and that the threshold energy was based on the #"0.57 nm"# wavelength, which is a possible but nonphysical mistake.


We first consider the photoelectric effect.

http://www.cyberphysics.co.uk/

As we are bombarding a metal surface to eject an electron, it has leftover kinetic energy #K_e# equal to the difference between the incoming energy #hnu# and the threshold energy #phi = hnu_0# that was overcome:

#K_e = hnu - hnu_0 = hnu - hnu_0#

where #h# is Planck's constant.

Recall that for every wavelength #lambda# that light, with speed #c = 2.998xx10^(8) "m/s"#, cycles through, we speak of a frequency #nu#. Thus, #nu = c/lambda#, and:

#K_e = (hc)/(lambda) - hnu_0#

Since the photoelectrons are to have the wavelength of #"0.57 nm"#, having mass, they follow the de Broglie relation:

#lambda = h/(mv) = h/p#

where #m#, #v#, and #p# are mass, speed, and momentum, respectively.

Recall that for mass-ive particles, #K_e = 1/2mv^2 = p^2/(2m)#. Thus,

#K_e = (h//lambda)^2/(2m)#

#= (6.626 xx 10^(-34) "kg"cdot"m"^2cdot"s"^(-1) // 0.57 xx 10^(-9) "m")^2/(2 xx 9.109 xx 10^(-31) "kg")#

#= 7.417 xx 10^(-19) "J"#,

or #"4.63 eV"#, is the energy of the photoelectrons leaving potassium metal. The actual ionization energy of potassium is #"4.34 eV"#, which means that based on that threshold energy, we have:

#overbrace("4.63 eV")^(K_e) = (hc)/lambda - overbrace("4.34 eV")^(hnu_0)#

or

#overbrace(7.417 xx 10^(-19) "J")^(K_e) = (hc)/lambda - overbrace(6.952 xx 10^(-19) "J")^(hnu_0)#

Thus,

#color(blue)(lambda) = (hc)/((7.417 + 6.952) xx 10^(-19) "J")#

#= (6.626 xx 10^(-34) "J"cdot"s" cdot 2.998 xx 10^(8) "m/s")/(1.437 xx 10^(-18) "J")#

#= 1.382 xx 10^(-7) "m"#

#=# #color(blue)("138.24 nm")#