Chapter-12-HW

Chapter-12-HW - Homework #5 Chapter 12 Quantum Mechanics...

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1 Homework #5 Chapter 12 Quantum Mechanics and Atomic Theory 13. Z is the atomic number or number of protons. Z eff is the effective nuclear charge or the nuclear charge after taking into account the shielding caused by other electrons in the atom. 14. The ionization energy is the energy it takes to remove an electron from an atom. As one goes from left to right across the periodic table in 1 row, electrons are added to the same shell. As each electron is added a proton is also added to the nucleus. Because the electrons are added to the same shell the shielding of these electrons on the nucleus is minimal. This causes the effective nuclear charge to go up. The larger the positive effective nuclear charge the tighter the electrons are held to the nucleus causing a greater amount of energy to be needed to remove an electron. As you go down a period you are adding electrons to a new shell. Each new shell causes substantial shielding of the nucleus. In addition, the electrons are farther away from the nucleus causing less energy to be needed to remove a valence election. 21. c    1 100 8 10 1.0 0.010 3.00 10 3.0 10 0.010 m cm m s cm m c Hz m  Energy of 1 photon    34 10 23 6.626 10 3.0 10 2.0 10 E h J s Hz J  Energy of avogadros photons    mol J mol A J E N 12 10 0 . 2 022 . 6 23 1 22. Wave a has a longer wavelength than wave b . Wave b has the higher frequency than wave a . Wave b has the higher energy than wave a . In 1.6×10 -3 m wave a completes 4 cycles. m m 4 3 10 0 . 4 4 10 6 . 1 Hz m c c s m 11 4 8 10 5 . 7 10 0 . 4 10 998 . 2    34 11 22 6.626 10 7.5 10 4.9 10 E h J s Hz J In 1.6×10 -3 m wave b completes 8 cycles. m m 4 3 10 0 . 2 8 10 6 . 1
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2 Hz m c c s m 12 4 8 10 5 . 1 10 0 . 2 10 998 . 2     34 12 22 6.626 10 1.5 10 9.9 10 E h J s Hz J  All electromagnetic radiation travels at the same velocity, the speed of light. Both waves a and b are infrared radiation. 23. a) c 8 6 13 2.998 10 5.0 10 6.0 10 m s c m Hz b) Infrared c)    34 13 20 6.626 10 6.0 10 4.0 10 E h J s Hz J This is the energy per photon, calculate the energy per mol of photons     mol J mol A J EN 4 1 23 20 10 4 . 2 022 . 6 0 . 4 d) This bond absorbs radiation with a smaller frequency; therefore, the radiation is less energetic. 25. Calculate the energy needed to remove 1 electron J N E mol mol J A 19 1 10 645 . 4 10 022 . 6 279700 Calculate the frequency of the photon needed 19 14 34 4.645 10 7.010 10 6.626 10 Eh EJ Hz h J s  Calculate the wavelength of the photon   9 8 7 10 1 14 2.998 10 4.277 10 427.7 7.010 10 m s nm m c c m nm Hz 26.
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Chapter-12-HW - Homework #5 Chapter 12 Quantum Mechanics...

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