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 energy of the n = 3 level
C)the sum of the energies of n = 2 and n = 3
D)the difference in energies between n = 2 and n = 3
E)all of these

Question

Quizplus · Accepted Answer

The information given in the problem allows us to determine the energy difference between the n=2 and n=3 levels, as this is the energy provided by the photon. We can calculate this energy difference using the equation:

∆E = E(n=3) - E(n=2) = 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 values given in the problem, we get:

∆E = (6.626 x 10^-34 Js)(3.00 x 10^8 m/s)/(500 x 10^-9 m) = 3.98 x 10^-19 J

Therefore, the answer is D, the difference in energies between n=2 and n=3. We cannot determine the individual energies of the n=2 or n=3 levels, nor their sum, without additional information.

In an Investigation of the Electronic Absorption Spectrum of a Particular