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LABRAKE CH301 Exam 3 Fall 2017
50090
Remember to refer to the Periodic Table handout that is separate from this exam copy. There are many conversion factors and physical constants available there.
NOTE: Please keep this exam copy intact (all pages still stapled - including this cover page). You must turn in ALL the mate- rials that were distributed. This means that you turn in your exam copy (name and signature included), bubble sheet, pe- riodic table handout, and all scratch paper. Please also have your UT ID card ready to show as well.
This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering.
001 10.0 points An electron is designated by the following quantum numbers:
n = 5 ℓ = 2 mℓ = 0
What is the electron’s orbital designation?
- 5 f
- 6 f
- 6 d
- 5 d correct
Explanation: If n=5, then the electron is in shell 5. If ℓ=2, then the electron is in subshell d. Thus, the electron’s orbital designation is 5d.
002 10.0 points A radio station transmits a signal with a wave- length of 3.26 meters. At what frequency is the radio station broadcasting?
- 920 MHz
- 2 × 107 MHz
- 92.0 MHz correct
- 0.920 MHz
Explanation:
ν = (^) λc
ν = 3 × 10 8 m s 3 .26 m = 9.^2 ×^10
(^7) Hz
- 2 × 107 Hz = 92.0 MHz
003 10.0 points Which set of quantum numbers is possible for an electron in the partially filled subshell found in the ground state of a manganese (Mn) atom?
- n = 3, ℓ = 2, mℓ = − 3 , ms = +^12
- n = 4, ℓ = 1, mℓ = +1, ms = +^12
- n = 3, ℓ = 2, mℓ = − 2 , ms = +^12 correct
- n = 3, ℓ = 3, mℓ = +2, ms = +^12
- n = 4, ℓ = 2, mℓ = +2, ms = +^12
Explanation: The partially filled subshell found in the ground state of a manganese (Mn) atom is the 3d subshell. This means n = 3, ℓ = 2. The possible values for mℓ = − 2 , − 1 , 0 , +1, +2.
004 10.0 points The 2s and 2p subshells are degenerate in energy for: I. a 1 electron system II. multi-electron systems III. they are not degenerate
- I and II only
- I only correct
- III only
- II only
Explanation: Subshells are degenerate in energy for 1 elec- tron systems only.
005 10.0 points An electron in a hydrogen atom absorbs a photon and raises from the n = 1 to the n = 5 energy level. What is the energy of the photon absorbed?
- 05 × 10 −^16 J
- 09 × 10 −^16 J
- 05 × 107 J
- 09 × 10 −^18 J correct
are paramagnetic. Atoms with no unpaired electrons are diamagnetic.
010 10.0 points Which of the following statements is correct?
- The electron in a hydrogen atom can emit photons of any energy.
- The quantum mechanical approach to atomic structure permits calculating a region about the nucleus in which an electron of a particular energy is most likely to be found. correct
- It is not possible for an electron to be ejected from an atom.
- Black lines in an atomic absorption spec- tra represent the unique wavelength of pho- tons that are emitted when electrons in a sub- stance are excited to a higher energy level.
Explanation: The electron in a hydrogen atom (or any atom for that matter) can only emit pho- tons of discreet energy levels. Black lines in an atomic absorption spectra represent the unique wavelength of photons that are ab- sorbed when electrons in a substance are ex- cited to a higher energy level. The quantum mechanical approach to atomic structure per- mits calculating a region about the nucleus in which an electron of a particular energy is most likely to be found. It is possible for an electron to absorb so much energy that it is actually ripped off of the atom.
011 10.0 points What is the ground-state electron configura- tion expected for antimony (Sb)?
- [Kr] 5s^24 d^85 p^5
- [Kr] 5s^24 f 14 d^75 p^5
- [Kr] 5s^24 f 105 p^3
- [Kr] 5s^24 f 14 d^85 p^4
- [Kr] 5s^24 d^105 p^3 correct
- [Kr] 5s^24 d^95 p^4
- [Kr] 5s^24 f 24 d^65 p^5 Explanation: The Aufbau order of electron filling is 1s, 2s, 2 p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f , 5d, 6 p, etc. s orbitals can hold 2 electrons, p orbitals 6 electrons, and d orbitals 10 electrons. Finally use noble gas shorthand to get the answer: [Kr] 5s^24 d^105 p^3.
012 10.0 points You shine a 600 nm light at a metal surface and electrons are ejected. If you maintain the same wavelength but increase the intensity of the light, what do you expect would happen? I. Ejected electrons would have a higher KE. II. Ejected electrons would have a lower KE. III. More electrons would be ejected. IV. Less electrons would be ejected.
- IV only
- I only
- II and IV only
- II only
- I and III only
- III only correct Explanation: Increasing the intensity of light does not change the amount of energy per photon. Rather, it only increases the number of pho- tons released per unit time. Thus, the KE of ejected electrons would remain the same, but more electrons would be ejected.
013 10.0 points In the experimental set-up for the photoelec- tric effect, the incident photons have an en-
ergy of 8.0 eV per photon. The work function of the metal surface is 1.0 eV. Which of the following statements is true?
I. Two electrons can be ejected per photon, each with a kinetic energy of 3.0 eV.
II. One electron can be ejected per photon with a kinetic energy of 7.0 eV.
III. No electrons can be ejected from the surface.
IV. One electron can be ejected per photon with a kinetic energy of 1.0 eV.
- I and IV only
- III only
- I and II only
- II only correct
- IV only
- I only
Explanation: In this case, as the energy per photon is greater than the work function, electrons will be ejected. Only one electron can be ejected per photon. Any energy from the photon in excess of the work function becomes the ki- netic energy of the electron.
Ephoton − φ = KEelec
8 .0 eV − 1 .0 eV = 7.0 eV
