Question

The allowed energies of a simple atom are 0.00 eV, 3.59 eV , and 5.97 eV

Part A

What wavelengths appear in the atom's emission spectrum?

Enter your answers in ascending order separated by commas.

Part B

What wavelengths appear in the atom's absorption spectrum?

Enter your answers in ascending order separated by commas.

Answer 1

Emission and absorption are opposites. Emission refers the a decrease in the energy of the atom from a higher state to a lower state and vice versa for absorption.

In this case, there are 6 answers. 3 for the first, 3 for the second. They are complements of each other. I'll label each case of emission x and the case of absorption as x' for you to make better links.

Case 1) An electron transits from 3.59 eV to 0.0 eV, thereby emitting a photon.

E = hc / ?

3.59 * 1.6 * 10^-19 = (6.63*10^-34)(3*10^8) / ?

? = 3.462 * 10^-7 = 346.2 nm (UV line observed on spectra)

Case 1') An electron transits from 0.0 eV to 3.59 eV, thereby absorbing a photon.

E = hc / ?

3.59 * 1.6 * 10^-19 = (6.63*10^-34)(3*10^8) / ?

? = 3.462 * 10^-7 = 346.2 nm (dark line observed on spectra)

Case 2) An electron transits from 5.97 eV to 0.0 eV, thereby emitting a photon.

E = hc / ?

5.97 * 1.6 * 10^-19 = (6.63*10^-34)(3*10^8) / ?

? = 2.082 * 10^-7 = 208.2 nm (UV line observed on spectra)

Case 2') An electron transits from 0.0 eV to 5.97 eV, thereby absorbing a photon.

E = hc / ?

5.97 * 1.6 * 10^-19 = (6.63*10^-34)(3*10^8) / ?

? = 2.082 * 10^-7 = 208.2 nm (dark line observed on spectra)

Case 3) An electron transits from 5.97 eV to 3.59 eV, thereby emitting a photon.

E = hc / ?

2.38 * 1.6 * 10^-19 = (6.63*10^-34)(3*10^8) / ?

? = 5.223 * 10^-7 = 522.3 nm (Red line observed on spectra)

Case 3') An electron transits from 3.59 eV to 5.97eV, thereby absorbing a photon.

E = hc / ?

2.38 * 1.6 * 10^-19 = (6.63*10^-34)(3*10^8) / ?

? = 5.223 * 10^-7 = 522.3 nm (dark line observed on red-end of spectra)