Question

The work function of an element is the energy required to remove an electron from the surface of the solid element. The work function for lithium is 279.7 kJ/mol. What is the maximum wavelength of light that can remove an electron from an atom ?

Answer 1

The maximum wavelength of light that can remove an electron from a lithium atom is equal to `4.279 * 10^(-7)"m"`.

So, you know that the work function of lithium, which is the energy needed to remove an electron from an atom located at the surface of the metal, is equal to 279.7 kJ/mol.

In order to be able to calculate the energy required to remove a single electron from the surface of the metal, you need to convert the work force from kJ per mole to kJ per electron.

Since 1 mole of electrons is defined as containing `6.022 * 10^(23)` electrons - this is know as Avogadro's number, you can get the energy needed to remove 1 electron by dividing the work function by the number of electrons present in a mole

`279.7cancel("kJ")/cancel("mol") * (1cancel("mole"))/(6.022 * 10^(23)e^(-)) * "1000 J"/(1cancel("kJ")) = 4.645 * 10^(-19)"J"`

So, in order to remove 1 electron from the surface of lithium, you need to provide it with that much energy.

The energy of a photon can be written as

`E_"photon" = h * nu`, where

`h` - Planck's constant, equal to `6.626 * 10^(-34)"J s"`;
`nu` - the frequency of the light;

Since you need to determine the wavelength of the light, you can use the relationship that exists between frequency and wavelength to express the energy of the photon

`c = lamda * nu => nu = c/(lamda)`, where

`c` - the speed of light, equal to `3 * 10^(8)"m/s"`
`lamda` - the wavelength of the light.

This will get you

`E_"photon" = h * c/(lamda) => lamda = (h * c)/E_"photon"`

`lamda = (6.626 * 10^(-34)cancel("J") * cancel("s") * 3 * 10^(8)"m" * cancel("s"^(-1)))/(4.645 * 10^(-19)cancel("J")`

`lamda = color(green)(4.279 * 10^(-7)"m")` `->` rounded to four sig figs