Calculate the energy of the following electronic transition and determine whether the radiation is emitted or absorbed from n=2 to n=6

To determine the energy of the electronic transition, we'll use the formula: ΔE = -R * (1/n^2_final - 1/n^2_initial) Where: R is the Rydberg constant (approximately 2.18 x 10^-18 Joules). For transition from n=2 to n=6: n_initial = 2 n_final = 6 Plugging in the values: ΔE = -2.18 x 10^-18 * (1/6^2 - 1/2^2) ΔE = -2.18 x 10^-18 * (1/36 - 1/4) ΔE = -2.18 x 10^-18 * (-0.2222) ΔE = 4.84 x 10^-18 Joules Since ΔE is positive, energy is absorbed for the transition from n=2 to n=6.