# Question #79130

##### 1 Answer

#### 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 thekinetic energyof the ejected electron#E_"photon"# is theenergyof the photon#W# is thework functionof 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 theenergyof the photon#h# isPlanck's constant, equal to#6.626 * 10^(-34)color(white)(.)"J s"# #c# is thespeed of lightin 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 **mass** of the electron.

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)#