Chapter 6, Problem 6.105MP

The ionization energy of an atom can be measured by photo electron spectroscopy, in which light of wavelength $\lambda$ is directed at an atom, causing an electron to be ejected. The kinetic energy of the ejected electron $\left(E_K\right)$ is measured by determining its velocity, v since $E_K=1\text{/}2m{v}^2$ . The $E_i$ is then calculated using the relationship that the energy of the incident light equals the sum of $E_i$ plus $E_K$ .
(a) What is the ionization energy of rubidium atoms in kilojoules per mole if light with $\lambda =58.4\text{nm}$ produces electrons with a velocity of $2.450\times {10}^6\text{m/s}$ ? (The mass ofan electron is $9.109\times {10}^{-31}\text{kg}$ .)
(b) What is the ionization energy of potassium in kilojoulesper mole if light with $\lambda =142\text{nm}$ produces electrons witha velocity of $\text{1}{\text{.240×10}}^6\text{m/s}$ ?