Problem 3824 PartA The allowed energies of a simple atom are

Problem 38.24 PartA The allowed energies of a simple atom are 0.00 eV, 4.42 eV and 6.04 eV What wavelengths appear in the atom\'s emission spectrum? Enter your answers in ascending order separated by commas. nm Submit Mx Answers Give Up Incorrect: Try Again; 6 attempts remaining; no points deducted Your answer does not have the correct number of comma-separated terms rt Part B What wavelengths appear in the atom\'s absorption spectrum? Enter your answers in ascending order separated by commas. nin Submit My Answers Give Up

Solution

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.

Case 1) An electron transits from 4.420 eV to 0.0 eV, thereby emitting a photon.

E = hc /

4.42 * 1.6 * 10^-19 = (6.63*10^-34)(3*10^8) /

= 2.8125 * 10^-7 = 281.25 nm (UV line observed on spectra)

Case 1\') An electron transits from 0.0 eV to 4.40 eV, thereby absorbing a photon.

E = hc /

4.42 * 1.6 * 10^-19 = (6.63*10^-34)(3*10^8) /

= 2.8125 * 10^-7 = 281.25 nm (dark line observed on spectra)

Case 2) An electron transits from 6.04 eV to 0.0 eV, thereby emitting a photon.

E = hc /

6.04 * 1.6 * 10^-19 = (6.63*10^-34)(3*10^8) /

= 2.058 * 10^-7 = 205.8 nm (UV line observed on spectra)

Case 2\') An electron transits from 0.0 eV to 6.04 eV, thereby absorbing a photon.

E = hc /

6.04 * 1.6 * 10^-19 = (6.63*10^-34)(3*10^8) /

= 2.058 * 10^-7 = 205.8 nm (dark line observed on spectra)

Case 3) An electron transits from 6.04 eV to 4.42 eV, thereby emitting a photon.

E = hc /

1.62 * 1.6 * 10^-19 = (6.63*10^-34)(3*10^8) /

= 7.674 * 10^-7 = 767.4 nm (Red line observed on spectra)

Case 3\') An electron transits from 4.0 eV to 6.0 eV, thereby absorbing a photon.

E = hc /

1.62 * 1.6 * 10^-19 = (6.63*10^-34)(3*10^8) /

= 7.674 * 10^-7 = 767.4 nm (dark line observed on red-end of spectra)