In an investigation of the electronic absorption spectrum of a particular element, it is found that a photon having λ = 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

The energy difference between the n = 2 and n = 3 levels can be calculated using the formula ΔE = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon. Plugging in the given values, we get ΔE = (6.626 x 10^-34 J s)(3.0 x 10^8 m/s)/(500 x 10^-9 m) = 3.98 x 10^-19 J. This is the energy required to promote the electron from the n = 2 to n = 3 level.

In an Investigation of the Electronic Absorption Spectrum of a Particular