Question #79130
1 Answer
Here's what I got.
Explanation:
The idea here is that when the incident photon strikes the surface of the metal, it transfers its energy to a surface electron, which gains a kinetic energy
However, this will only happen if the incoming photon has enough energy to overcome the work function of the metal.
#color(blue)(ul(color(black)("K"_ "E" = E_"photon" - W)))" " " "color(darkorange)(("*"))#
Here
#"K"_"E"# is the kinetic energy of the ejected electron#E_"photon"# is the energy of the photon#W# is the work function of the metal
Now, I'm not really sure if the question is written down properly because you don't need to know the speed of the ejected electron to figure out the maximum wavelength that can eject electrons from potassium.
The longest wavelength of the photon, i.e. its lowest frequency, and thus its lowest energy, that can eject an electron from potassium is equal to the work function of the metal.
In other words, a photon can eject an electron from the surface of the metal if its energy satisfies
#E_"photon" >= W#
So you can say that your photon needs a minimum energy equal to the work function of the metal in order for an electron to be ejected from the surface.
#E_"photon min" = "2.29 eV"#
Convert this value to joules
#2.29 color(red)(cancel(color(black)("eV"))) * (1.6 * 10^(-19)color(white)(.)"J")/(1color(red)(cancel(color(black)("eV")))) = 3.664 * 10^(-19)color(white)(.)"J"#
Now, the energy of the photon is directly proportional to its frequency, which implies that it's inversely proportional to its wavelength, as shown by the Planck - Einstein relation
#E = (h * c)/lamda# Here
#E# is the energy of the photon#h# is Planck's constant, equal to#6.626 * 10^(-34)color(white)(.)"J s"# #c# is the speed of light in a vacuum, usually given as#3 * 10^8color(white)(.)"m s"^(-1)#
Rearrange the equation to solve for the wavelength of the photon.
#lamda = (h *c)/E#
Plug in your value to find
#lamda = (6.626 * 10^(-34) color(red)(cancel(color(black)("J"))) color(red)(cancel(color(black)("s"))) * 3 * 10^(8) color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))))/(3.664 * 10^(-19)color(red)(cancel(color(black)("J"))))#
#color(darkgreen)(ul(color(black)(lamda = 5.43 * 10^(-7)color(white)(.)"m")))#
The answer is rounded to three sig figs.
You can thus say that in order for an electron to be ejected from the surface of potassium, the incoming photon must have a maximum wavelength of
If the wavelength of the incoming photon is longer than this value, then you won't see an emission of electrons from the surface of the metal.
Similarly, if the wavelength of the incoming electron is shorter than this value, then the kinetic energy of the ejected electrons will increase accordingly.
To find the wavelength of the photon that would give the ejected electron a speed of
#"668 km s"^(-1) = 6.68 * 10^5color(white)(.)"m s"^(-1)#
use the fact that the kinetic energy of the electrons is equal to
#"K"_"E" = 1/2 * m * v^2#
Here
Plug this into equation
#E_"photon" = 1/2 * m * v^2 - W#
This is equivalent to
#(h * c)/lamda_ "668 km/s" = 1/2 * m * v^2#
which gets you
#lamda_ "668 km/s" = (2 * h * c)/(m * v^2)#
