The ionization energy of an atom can be measured by photoelectron spectroscopy, in which light of wavelength is directed at an atom, causing an electron to be ejected. The kinetic energy of the ejected electron(Ek) is measured by determining its velocity,v, since Ek=1/2mv^2. The Ei is then calculated using the relationship that the energy of the incident light equals the sum of Ei plus Ek
1) What is the ionization energy of rubidium atoms (in kilojoules per mole) if light with Lambda = 58.4 nm produces electrons with a velocity of 2.450 x 10^6 m/s? (The mass of an electron is 9.109 x 10^-31 kg)
2) What is the ionization energy of potassium (in kilojoules per mole) if light with Lambda = 142 nm produces electrons with a velocity of 1.240 x 10^6 m/s?