ism_chapter_36

ism_chapter_36 - Chapter 36 Atoms Conceptual Problems *1...

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1173 Chapter 36 Atoms Conceptual Problems *1 Determine the Concept Examination of Figure 35-4 indicates that as n increases, the spacing of adjacent energy levels decreases . 2 Picture the Problem The energy of an atom of atomic number Z , with exactly one electron in its n th energy state is given by ... 3, 2, 1, , 2 2 0 2 = = n n E Z E n . Express the energy of an atom of atomic number Z , with exactly one electron, in its n th energy state: ... 3, 2, 1, , 2 2 0 2 = = n n E Z E n where E 0 is the atom’s ground state energy. For lithium ( Z = 3) in its first excited state ( n = 1) this expression becomes: () 0 2 0 2 2 9 1 3 E E E = = and correct. is ) ( a 3 Determine the Concept Bohr’s postulates are 1) the electron in the hydrogen atom can move only in certain non-radiating, circular orbits called stationary states , 2) if E i and E f are the initial and final energies of the atom, the frequency f of the emitted radiation during a transition is given by f = [ E i E f ]/ h , and 3) the angular momentum of a circular orbit is constrained by mvr = n h . correct. is ) ( a 4 •• Picture the Problem We can express the kinetic energy of the orbiting electron as well as its total energy as functions of its radius r . Express the total energy of an orbiting electron: U K E + = Express the orbital kinetic energy of an electron: r kZe K 2 2 = (1) Express the potential energy of an orbiting electron: r kZe U 2 = Substitute and simplify to obtain: r kZe r kZe r kZe r kZe r kZe E 2 2 2 2 2 2 2 2 2 2 = = =
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Chapter 36 1174 . therefore and negative less becomes increases, as Thus, increases E r decreases. increases, if clear that it makes for expression the of n Examinatio K r K 5 Picture the Problem We can relate the kinetic energy of the electron in the n = 2 state to its total energy using 2 2 2 U K E + = . Express the total energy of the hydrogen atom in its n = 2 state: 2 2 2 2 2 2 2 K K K U K E = = + = or 2 2 E K = Express the energy of hydrogen in its n th energy state: () 2 2 0 2 2 0 2 2 2 0 2 1 n E n E n E Z E n = = = where E 0 is hydrogen’s ground state energy and Z = 1. Substitute to obtain: 2 2 0 n E K n = and 4 2 0 2 0 2 E E K = = correct. is ) ( d 6 Picture the Problem The orbital radius r depends on the n = 1 orbital radius a 0 , the atomic number Z , and the orbital quantum number n according to r = n 2 a 0 / Z . The radius of the n = 5 orbit is: 0 0 2 5 25 1 5 a a r = = because Z = 1 for hydrogen. ( ) correct. is b *7 Determine the Concept We can find the possible values of l by using the constraints on the quantum numbers n and l . The allowed values for the orbital quantum number l for n = 1, 2, 3, and 4 are summarized in table shown to the right: n l 10 20 , 1 3 0, 1, 2 4 0, 1, 2, 3 From the table it is clear that l can have 4 values. correct. is ) ( a
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Atoms 1175 8 Picture the Problem We can find the number of different values m l can have by enumerating the possibilities when the principal quantum number n = 4.
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ism_chapter_36 - Chapter 36 Atoms Conceptual Problems *1...

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