Question

In: Physics

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

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.

Solutions

Expert Solution

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)


Related Solutions

The allowed energies of a simple atom are 0.0 eV, 4.5 eV and 7.0 eV ....
The allowed energies of a simple atom are 0.0 eV, 4.5 eV and 7.0 eV . An electron traveling at a speed of 1.8×106 m/s collisionally excites the atom. What is the minimum speed the electron could have after the collision? What is the maximum speed the electron could have after the collision?
Electrons of energies 10.2 eV, 12.09 eV, and 13.06 eV colliding with a hydrogen atom, can...
Electrons of energies 10.2 eV, 12.09 eV, and 13.06 eV colliding with a hydrogen atom, can cause radiation to be emitted from the latter. Calculate in each case the principle quantum number, n, of the orbit to which the electron in the hydrogen atom is raised and the wavelength of the radiation emitted if it drops back to the ground state.
A certain quantum system has allowed energies of 1.0 eV, 2.0 eV, 4.0 eV, and 7.0...
A certain quantum system has allowed energies of 1.0 eV, 2.0 eV, 4.0 eV, and 7.0 eV. What wavelengths appear in the system’s emission spectrum?
The orbital potential energies for the 4s and 4p of Br are -18.65 ev and -12.49ev,...
The orbital potential energies for the 4s and 4p of Br are -18.65 ev and -12.49ev, respectively. Recall that the H 1s is -13.61ev. The MO diagram for HBr is not exactly the same as HF. Draw the MO diagram for HBr, including energy levels, each orbitals shape, each orbitals character (meaning what atomic orbitals contribute to each M.O. and how much), symmetry labels (e.g.  and , * and *), indicate the HOMO and LUMO, the bond order (show...
For the following reaction,  10.1 grams of  nitrogen gas are allowed to react with  5.97 grams of  oxygen gas ....
For the following reaction,  10.1 grams of  nitrogen gas are allowed to react with  5.97 grams of  oxygen gas . nitrogen(g) +  oxygen(g)   nitrogen monoxide(g) What is the maximum mass of  nitrogen monoxide that can be formed?   grams What is the  FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete?   grams
. Given that the activation energy for vacancy formation in Copper is 0.90 eV/atom, calculate the...
. Given that the activation energy for vacancy formation in Copper is 0.90 eV/atom, calculate the temperature (in Kelvin) at which 1 of every 10,000 Copper lattice sites is vacant.
Consider an atom with four energy levels, where the lowest energy is -10 eV and the...
Consider an atom with four energy levels, where the lowest energy is -10 eV and the highest level is 0 eV. Sketch an energy-level diagram that could produce the spectrum. What are the energies of the other two energy levels. (Note: there are more than two answers.) A -2 eV and -5 eV B -1 eV and -4 eV C -1 eV and -6 eV D -6 eV and -9 eV E -7 eV and -3 eV
Assume an atom has K, L, and M electrons and the energies for K, L, and...
Assume an atom has K, L, and M electrons and the energies for K, L, and M shells are -5000, -500, and -50 eV. Please give all possible energies of Auger electrons and characteristic X-rays.
The first five ionization energies of an atom X are 6, 19, 28,120, and 154...
The first five ionization energies of an atom X are 6, 19, 28, 120, and 154 eV. Sugust what 3rd row element X is and give reasons for your choice.
The energies of the hydrogen atom are quantized by integer number n from 1 to infinity....
The energies of the hydrogen atom are quantized by integer number n from 1 to infinity. This number if called the principal quantum number. Find: a) principal quantum number b) binding energy c) orbital radius d) total enery e) excitation energy of the fifth state of hydrogen
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT