It takes \(208.4 \mathrm{~kJ}\) of energy to remove \(1 \mathrm{~mole}\) of electrons from an atom on the surface of rubidium metal. How much energy does it take to remove a single electron from an atom on the surface of solid rubidium? What is the maximum wavelength of light capable of doing this?

First, we have to convert the given energy from kJ to J to make the calculations easier. 1 kJ = 1000 J. So, 208.4 kJ/mol = 208.4 * 1000 J/mol = 208400 J/mol

Avogadro's number = 6.022 * 10^23 (number of atoms in 1 mole) We will divide the energy required to remove 1 mole of electrons (208,400 J/mol) by Avogadro's number to find the energy that is required to remove one electron. Energy per electron = \( \frac{208,400 \mathrm{~J/mol}}{6.022 \times 10^{23} / \mathrm{mol}}\) Energy per electron = \(3.461 \times 10^{-19} \mathrm{J}\)

The relationship between the energy of a photon (E) and its wavelength (\(λ\)) is given by: E = \( \frac{hc}{λ}\), Where h = Planck's constant = \(6.626 \times 10^{-34} \mathrm{Js}\), c = speed of light = \(3 \times 10^8 \mathrm{m/s}\) Rearrange the equation to find the wavelength: \(λ = \frac{hc}{E}\) Plug in the values for h, c, and energy per electron (E = \(3.461 \times 10^{-19} \mathrm{J}\)): \(λ = \frac{(6.626 \times 10^{-34} \mathrm{Js})(3 \times 10^8 \mathrm{m/s})}{3.461 \times 10^{-19} \mathrm{J}}\) Calculate the wavelength: \(λ = 5.728 \times 10^{-7} \mathrm{m}\) As the wavelength of light can be expressed in nanometers (nm), we can convert the calculated value from meters to nanometers: 1 meter = \(1 \times 10^9\) nanometers \(λ = 5.728 \times 10^{-7} \mathrm{m} * \frac{1 \times 10^9 \mathrm{nm}}{1 \mathrm{m}}\) \(λ = 572.8 \mathrm{nm}\) Thus, the energy required to remove one electron from an atom of rubidium on the surface is \(3.461 \times 10^{-19} \mathrm{J}\), and the maximum wavelength of light capable of doing this is 572.8 nm.