# If #6000# angstrom light is shined onto a metal work a certain work function, what is the wavelength in #"m"# of the fastest photoelectron that can be emitted?

##### 1 Answer

Here's what I got.

#### Explanation:

**!! VERY LONG ANSWER !!**

The idea here is that the metal has something called a **work function**, which basically represents the amount of energy needed in order to remove an electron from the surface of the metal.

Now, it's important to realize that not all the electrons emitted from the surface will have the same kinetic energy.

That happens because not all the energy carried by a photon will be transferred to the *surface electrons*. Some of this energy will actually be transferred to the bulk of the metal instead, i.e. to electrons that are **not** located near the surface.

This means that the **fastest electrons** emitted from the metal will absorb **all the energy** of an incoming photon **that is not** needed to remove the electron from the surface and **that is not transferred** to the bulk of the metal.

In other words, the maximum kinetic energy of an emitted electron is given by

#K_ "E max" = E_"photon" - W#

Here

#E_"photon"# represents the energy of the incoming photon#W# is thework functionof the metal

It's worth mentioning that if **photoelectric effect** does not take place!

Also, notice that the **slowest electron** emitted from the surface of the metal has

#K_"E min" ~~ "0 J"#

because, for all intended purposes, **all the energy** of the incoming photon is used up to overcome the work function, i.e.

Now, the **energy** of the photon is calculated by using the **Planck - Einstein equation**, which looks like this

#color(blue)(ul(color(black)(E = h * c/(lamda))))#

Here

#E# is theenergyof the photon#lamda# is the wavelength of the photon#c# is the speed of light in a vacuum, usually given as#3 * 10^8"m s"^(-1)# #h# isPlanck's constant, equal to#6.626 * 10^(-34)"J s"#

In your case, the wavelength is given in *angstrom*, so convert it to *meters* by using

#1color(white)(.)stackrel(@)("A") = 1 * 10^(-10)# #"m"#

You will end up with

#6000 color(red)(cancel(color(black)(stackrel(@)("A")))) * (1 * 10^(-10)color(white)(.)"m")/(1color(red)(cancel(color(black)(stackrel(@)("A"))))) = 6.0 * 10^(-7)# #"m"#

Plug this into the Planck - Einstein equation and find the energy of the incoming photon

#E = 6.626 * 10^(-34)"J"color(red)(cancel(color(black)("s"))) * (3 * 10^8 color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))))/(6.0 * 10^(-7)color(red)(cancel(color(black)("m"))))#

#E = 3.313 * 10^(-19)# #"J"#

Now, assuming that the **work function** of the metal is equal to

#K_ "E max" = 3.313 * 10^(-19)color(white)(.)"J" - Wcolor(white)(.)"J"#

#K_ "E max" = (3.313 * 10^(-19) - W)# #"J"#

#color(red)(!)# Keep in mind that the work functionmustbe expressed injoules per electronin order for the above equation to work, so if the problems provides the work function inkilojoules per mole, make sure to convert it before using it!

So, you now know that the fastest electron emitted from the surface of the metal has a kinetic energy equal to

To find the **de Broglie wavelength**, you must use the following equation

#color(blue)(ul(color(black)(lamda_ "matter" = h/p))) -># thede Broglie wavelength

Here

#p# is the momentum of the electron#lamda_ "matter"# is its de Broglie wavelength

As you know, the **momentum** of the electron depends on its **velocity**, **mass**,

#color(blue)(ul(color(black)(p = m * v)))#

This means that the de Broglie wavelength is equal to

#lamda_ "matter" = h/(m * v)#

Now, the kinetic energy of the fastest electron is defined as

#K_"E max" = 1/2 * m * v^2#

Rearrange to find the velocity of the electron

#v = sqrt( (2 * K_ "E max")/m)#

Use this expression for the velocity of the electron to find its de Broglie wavelength

#lamda_ "matter" = h/(m * sqrt( (2 * K_ "E max")/m))#

If you take the mass of the electron to be approximately equal to

#m_ ("e"^(-)) ~~ 9.10938 * 10^(-31)# #"kg"#

and use the fact that

#"1 J" = 1# #"kg m"^2"s"^(-2)#

you can say that the de Broglie wavelength will be equal to

#lamda_ "matter" = (6.626 * 10^(-34) color(blue)(cancel(color(black)("kg"))) "m"^color(purple)(cancel(color(black)(2)))color(green)(cancel(color(black)("s"^(-2)))) * color(green)(cancel(color(black)("s"))))/(9.10938 * 10^(-31)color(blue)(cancel(color(black)("kg"))) * sqrt( (2 * (3.313 * 10^(-19) - W) color(red)(cancel(color(black)("kg"))) color(purple)(cancel(color(black)("m"^2)))color(green)(cancel(color(black)("s"^(-2)))))/(9.10938 * 10^(-31)color(red)(cancel(color(black)("kg")))))#

which gets you

#lamda_ "matter" = (7.2738 * 10^(-4))/(1.4817 * 10^(15) * sqrt((3.313 * 10^(-19) - W))# #"m"#

#lamda_"matter" = (4.91 * 10^(-19))/(sqrt((3.313 * 10^(-19)-W))# #"m"#

At this point, all you have to do is plug in the value you have for the work function in *joules per electron* and find the value of