An argon-ion laser is used in some laser light shows. The argon ion has strong emissions at \(485 \mathrm{~nm}\) and \(512 \mathrm{~nm}\). (a) What is the color of these emissions? (b) What is the energy associated with these emissions in kilojoules per mole? (c) Write the ground state electron configuration and orbital diagram of \(\mathrm{Ar}^{+}\).

The given wavelengths are 485 nm and 512 nm. Generally, - violet: 380-450 nm - blue: 450-495 nm - green: 495-570 nm - yellow: 570-590 nm - orange: 590-620 nm - red: 620-750 nm Since 485 nm is between 450-495 nm, the color corresponding to this wavelength is blue. Similarly, 512 nm lies in the green range (495-570 nm), so its color is green.

We'll use the Planck's equation to find the energy of the emitted light: \(E = \dfrac{hc}{\lambda}\) Where: \(E\) is the energy of the light \(h\) is the Planck's constant (\(6.626 \times 10^{-34} \mathrm{Js}\)) \(c\) is the speed of light (\(2.998 \times 10^8 \mathrm{m/s}\)) \(\lambda\) is the wavelength of the light We need to first convert the wavelength into meters: a) \(\lambda_1 = 485 \times 10^{-9} \mathrm{m}\) b) \(\lambda_2 = 512 \times 10^{-9} \mathrm{m}\) Now, we'll calculate the energy, E, in Joules: a) \(E_1 = \dfrac{(6.626 \times 10^{-34} \mathrm{Js})(2.998 \times 10^8 \mathrm{m/s})}{(485 \times 10^{-9} \mathrm{m})} = 4.089 \times 10^{-19} \mathrm{J}\) b) \(E_2 = \dfrac{(6.626 \times 10^{-34} \mathrm{Js})(2.998 \times 10^8 \mathrm{m/s})}{(512 \times 10^{-9} \mathrm{m})} = 3.895 \times 10^{-19} \mathrm{J}\) Next, we will convert the energy in Joules to kilojoules per mole: a) \((4.089 \times 10^{-19} \mathrm{J})(\dfrac{1 \mathrm{kJ}}{10^3 \mathrm{J}})(\dfrac{6.022 \times 10^{23} \mathrm{atoms/mole}}{1}) = 246.2 \mathrm{kJ/mol}\) b) \((3.895 \times 10^{-19} \mathrm{J})(\dfrac{1 \mathrm{kJ}}{10^3 \mathrm{J}})(\dfrac{6.022 \times 10^{23} \mathrm{atoms/mole}}{1}) = 234.5 \mathrm{kJ/mol}\) Thus, the energy of the emissions at 485 nm and 512 nm are 246.2 kJ/mol and 234.5 kJ/mol, respectively.

The neutral argon atom has 18 electrons, with the electron configuration: \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6}\) When an electron is removed to form the Ar+ ion, it has 17 electrons. Thus, the electron configuration of Ar+ will be: \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^5}\) The orbital diagram for Ar+ can be represented as follows: 1s: ↑↓ 2s: ↑↓ 2p: ↑↓ ↑↓ ↑↓ 3s: ↑↓ 3p: ↑↓ ↑↓ ↑ The ground state electron configuration and orbital diagram of Ar+ are \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^5}\) and the diagram represented above, respectively.