In an investigation of the electronic absorption spectrum of a particular element, it is found that a photon having $\lambda$ = 500 nm provides just enough energy to promote an electron from the second quantum level to the third. From this information, we can deduce
A) the energy of the n = 2 level.
B) the sum of the energies of the n = 2 and n = 3 levels.
C) the difference between the energies of the n = 2 and n = 3 levels.
D) the energy of the n = 3 level.
E) all of these.

Question

Quizplus · Accepted Answer

By using the formula E = hc/λ, where E is energy, h is Planck's constant, c is the speed of light, and λ is wavelength, we can find the energy of the photon with wavelength 500 nm. Then, we can calculate the energy difference between the n = 2 and n = 3 levels by using the formula ΔE = E_final - E_initial = E_3 - E_2, where ΔE is the energy difference, and E_final and E_initial are the energies of the final and initial states, respectively. Therefore, we can deduce the energy difference between the n = 2 and n = 3 levels, which is choice C. We cannot determine the individual energies of either level, so choices A and D are incorrect. We also cannot calculate the sum of the energies of the n = 2 and n = 3 levels, so choice B is incorrect. Therefore, the best choice is C.

In an Investigation of the Electronic Absorption Spectrum of a Particular