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# 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 ?

The maximum 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## - , equal to ##6.626 * 10^(-34)"J s"##; ##nu## - the 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