Unformatted text preview: tum number. The radius is inversely proportional to Z.
Therefore, we expect that the radius calculated with Z = 11 will be less than the radius
calculated with Z = Zeffective, where Zeffective > 1, but less than 11. This makes sense, because
11 protons would pull the outermost electron in more tightly than 1 or 2 protons would. SOLUTION
a. According to the Bohr model, the total energy E of the electron is En = − (13.6 eV ) Z2
n2 (30.13) To remove an electron with this energy from an atom, it is necessary to give the electron
positive energy of a magnitude equal to that in Equation 30.13. This energy is the ionization
energy Eionization. Then, the electron will be raised from the nth level to the n = ∞ level,
where the total energy of the electron is zero and the electron is separated from the atom.
Thus, with Z = Zeffective and n = 3 for the outermost electron in a sodium atom
(1s2 2s2 2p6 3s1) we have Eionization = (13.6 eV ) 2
Z effective 32 or Zeffective = 32 Eionization
13.6 eV 32 ( 5.1 eV )
=
= 1.8
13.6 eV As expected, Zeffective is less than 11 and greater than 1.
b. Using Equation 30.10 for the radius with Z = 11 and Z = Zeffective = 1.8, we obtain
[Z = 11] ( r3 = 5.29 × 10−11 m 3
) 11 = 4.3 ×10−11 m
2 Chapter 30 Problems ( r3 = 5.29 ×10−11 m [Z = 1.8] ) 1555 32
= 2.6 ×10−10 m
1.8 As expected, the radius is smaller when Z = 11. 17. SSM REASONING A wavelength of 410.2 nm is emitted by the hydrogen atoms in a
highvoltage discharge tube. This transition lies in the visible region (380–750 nm) of the
hydrogen spectrum. Thus, we can conclude that the transition is in the Balmer series and,
therefore, that nf = 2. The value of ni can be found using Equation 30.14, according to which
the Balmer series transitions are given by
1
= R 12 2
2
λ () 1 1 ni2 n = 3, 4, 5, . . . This expression may be solved for ni for the energy transition that produces the given
wavelength. SOLUTION Solving for ni, we find that ni = 1
1
2 2 1 =
1
Rλ 1
2 2 =6
1
−1 (1.097 × 10 m )(410.2 × 10
7 −9 m) Therefore, the initial and final states are identified by n i = 6 and nf = 2 .
______________________________________________________________________________
18. REASONING The energy levels and radii of a hydrogenic species of atomic number Z are
given by Equations 30.13 and 30.10, respectively: En = –(13.6 eV)( Z 2 / n2 ) and
–11
2
rn = (5.29 × 10 m)( n /Z ) . We can use Equation 30.13 to find the value of Z for the
unidentified ionized atom and then calculate the radius of the n = 5 orbit using
Equation 30.10. SOLUTION Solving Equation 30.13 for atomic number Z of the unknown species, we have Z= En n 2
–13.6 eV = (–30.6 eV)(2)2
=3
–13.6 eV 1556 THE NATURE OF THE ATOM Therefore, the radius of the n = 5 orbit is 52 –10
r5 = (5.29 × 10 –11 m) m = 4.41 × 10
3
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19. SSM REASONING Singly ionized helium, He + , is a hydrogenlike species with Z = 2 .
The wavelengths of the series of lines produced when the electron makes a transition from
higher energy levels into the nf = 4 level are given by Equation 30.14 with Z = 2 and nf = 4: 1
= 1.097 × 107 m –1 (22 ) 2
4
λ 1 ( ) 1 ni2 SOLUTION Solving this expression for ni gives 1
ni = 2
4 7
–1 4λ (1.097 × 10 m ) 1 –1/ 2 Evaluating this expression at the limits of the range for λ, we find that n i = 19.88 for
λ = 380 nm, and n i = 5.58 for λ = 750 nm. Therefore, the values of ni for energy levels
from which the electron makes the transitions that yield wavelengths in the range between
380 nm and 750 nm are 6 ≤ ni ≤ 19 .
______________________________________________________________________________
20. REASONING For each species, the wavelengths appearing in the line spectra can be
calculated using Equation 30.14:
1
1
= RZ 2 2 – 2 λ
ni nf
1 The shortest wavelength for a given series of lines occurs when an electron is in an initial
energy level with a principal quantum number of ni = ∞. Therefore, 1/ni2 = 1/∞ = 0, and the
shortest wavelength is given by
1 RZ 2
=2
λ
nf
In this expression, the value of nf is the same for the series for Li2+ and for Be3+. The term
R is the Rydberg constant and is also the same for both ionic species. Thus, rearranging this
equation gives Chapter 30 Problems λZ =
2 nf2
R 1557 = constant which is the basis for our solution.
SOLUTION Since λ Z2 is the same for each species, we have ( λ Z )Li = ( λ Z )Be
2 2 or λBe ( λ Z 2 )Li = ( 40.5 nm )(3)2 =
=
( 4 )2 2
Z Be 22.8 nm ______________________________________________________________________________
21. REASONING AND SOLUTION The shortest wavelength, λs, occurs when ni = ∞ , so that
1/ni = 0. In that case Equation 30.14 becomes
RZ 2
=2
λs
nf
1 or RZ =
2 nf2 λs The longest wavelength in the series occurs when ni = nf + 1. 1
1 n2 1
1 = RZ 2 2 – 2 = f 2 –
n λl
ni λs nf ( n + 1)2 f f 1 A little algebra gives nf as follows: λ1 41.02 ×10−9 m (nf + 1) 2
=
=
λs 22.79 ×10−9 m 2nf + 1
Or
nf2 – 1.600 nf – 0.800 = 0
Solv...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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