Answer Key for the Practice Questions for Atomic Theory

1. a) 
2. 
   b) 

   c) This is in the visible spectrum.

   d)

3. Quantum theory only allows for discrete energies to be emitted, not energies at all wavelengths. The discrete energies come from the spacing of the energy levels in which the electrons in potassium move around. See the atomic theory additional information for more of an explanation.
4. 
5. 
   There are 2 unpaired electrons in each.
6. a) 1s²2s²2p¹ - correct   b) 1s²2s¹3s¹ excited   c) 1s²2s²2p⁶3s²2d² incorrect   d) 1s²2s⁴2p² incorrect   e) 1s¹2s¹excited
7. 
8. 
9. 
10. 
    The color of light associated with 688 nm is red.
11. 
12. 
13. 
    The color of light associated with 436 nm is violet.

Answer Key for the Atom and Atomic Theory Review Questions

1. D = m/V = 3.44 g/2.77 mL = 1.24 g/mL
2. V = m/D = 50.0 g/1.19 x 10^-3 g/cm³ = 4.2 x 10⁴ cm³.
3. For the four main ideas of Dalton's atomic theory, see your book, p. 84.
4. Electrons were discovered by J.J. Thomson using the cathode ray tube. See pp.86-88 for the complete description.
5. Rutherford shot alpha particles through thin gold foil. He expected all of the particles to go straight through. Most did, but some deflected. This lead him to two conclusions: first, that atoms are mostly empty space, and secondly, that there is a positively charged, dense nucleus present. See your book for more of a description.
6. a) Ba: p⁺=56, e^-= 56, n^o= 81   b) Cu: p⁺= 29, e^-= 29, n^o= 35   c) F: p⁺= 9, e^-= 9, n^o= 10
7. An isotope of an element contains the same number of protons and electrons but different number of neutrons.
8. 
9. 
10. J.J. Thomson's model was the "plum pudding" model - a positively charged blob with negative electrons stuck throughout it. Rutherford stated that there was a small, dense, positively charged nucleus with electrons outside the nucleus. Bohr also had a positive nucleus, but stated that the electrons were in definite orbits, or energy levels, around the nucleus. The quantum mechanical model also gives a positive nucleus, but states that the electrons are in not in a definite orbit, but rather in a probability region, or electron cloud.
11. An atomic orbital is the region of space around a nucleus where an electron will most likely be (probability region).
12. An s orbital has a spherical shape. It holds 2 electrons. A p orbital has a dumbbell shape. It also holds 2 electrons, however there are three p orbitals, one oriented along the x-axis, another along the y-axis, and the third along the z-axis.
13. See your text.
14. a) S 1s²2s²2p⁶3s²3p⁴, [Ne] 3s²3p⁴
15. 
    b) Rb 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s¹, [Kr] 5s¹
    c) Au 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d⁹, though this is an exception and actually ends in 6s¹4f¹⁴5d¹⁰, [Xe] 6s²4f¹⁴5d⁹
16. 
17. lowest wavelength: violet, indigo, blue, green, yellow, orange, red: highest wavelength
18. The electron in the hydrogen atom can be excited to various energy levels by absorbing energy. As it relaxes back to a lower state, it releases energy. By relaxing from different levels, different amounts of energy will be released, which produce a different color of light. See the atomic theory additional information for more of an explanation.
19. 
20. 
21. 
22. The ground state is the lowest energy level for an electron to occupy. An excited state is when an electron is above the lowest energy level available.
23. Heisenberg's uncertainty principle states that both the position and velocity of an electron cannot be known at the same time.
24. de Broglie hypothesized that all matter can exhibit wavelike properties. He derived an equation to describe the wavelength of a particle  where h is Planck's constant, m is the mass of the particle, and v is its velocity.
25. The photoelectric effect is where a photon (packet of light) can eject an electron from a metal. The photon must be at least at the threshold energy for an electron to be ejected. For a complete description, see your book.

Notes:

    E = energy  (Joules)
    f  = frequency (hertz)
    c = speed of light = 3.00 x 10⁸ m/s
    h = Planck's constant = 6.626 x 10^-34 J/s
    an upside down Y is lamda = wavelength (m)
    m = mass
    v  = velocity

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