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Unformatted text preview: ing the quadratic equation yields only one positive root, which is nf = 2. Therefore,
1/λs = 1/(22.79 × 10–9 m) = RZ2/nf2 = RZ2/4, which gives Z = 4. As a result, the nexttothelongest wavelength is:
1
1
= R(4)2 2 − 2 or
λ = 30.39 × 10−9 m = 30.39 nm
λ
4
2
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1 1558 THE NATURE OF THE ATOM 22. REASONING In either the ground state (n = 1) or the first excited state (n = 2), the
electron’s angular speed ωn is found from
vn = rnω n (8.9) where vn is the electron’s orbital speed, rn is the orbital radius, and ωn is measured in radians
per second. The speed vn of the electron depends upon the principal quantum number n via Ln = mvn rn = n h
2π n = 1, 2, 3, . .. (30.8) where m is the mass of the electron. The radius rn of the electron’s orbit is given by ( rn = 5.29 ×10−11 m ) nZ = (5.29 ×10−11 m) n2
2 n = 1, 2, 3, ... (30.10) where we have used Z = 1, the atomic number of hydrogen. After using Equations 8.9, 30.8,
and 30.10 to determine the angular speed ωn of the electron in radians per second, we will
convert the result to revolutions per second with the equivalence 1 revolution = 2π radians.
SOLUTION
a. Solving Equation 8.9 for ωn yields ωn = vn
rn (1) Solving Equation 30.8 for vn, we obtain vn = nh
2π m rn (2) Substituting Equation (2) into Equation (1), we find that nh vn 2π m rn nh =
ωn = =
2
rn
rn
2π m rn
Substituting Equation 30.10 into Equation (3) yields (3) Chapter 30 Problems ωn = nh
nh
=
2
2π m rn 2π m 5.29 ×10−11 m ( )( )
2 n 22 = h ( ) 2 2π m 5.29 ×10−11 m n3 1559 (4) To find the groundstate angular speed of the electron in revolutions per second, we
substitute n = 1 into Equation (4) and multiply by the appropriate conversion factor: ( 6.63×10−34 J ⋅ s) 1 rev 15
ω1 =
× = 6.59 ×10 rev/s
2
3 2π rad 2π ( 9.11×10−31 kg )( 5.29 ×10−11 m ) (1)
b. When the electron is in the first excited state, the value of the principal quantum number
is n = 2, so by Equation (4) we have that ( 6.63×10−34 J ⋅ s) 1 rev 14
ω2 =
× = 8.23 ×10 rev/s
2
3
2π ( 9.11×10−31 kg )( 5.29 ×10−11 m ) ( 2) 2π rad ______________________________________________________________________________
23. REASONING The orbital quantum number l has values of 0, 1, 2, ....., (n – 1), according
to the discussion in Section 30.5. Since l = 5, we can conclude, therefore, that n ≥ 6. This
knowledge about the principal quantum number n can be used with Equation 30.13,
En = –(13.6 eV)Z2/n2, to determine the smallest value for the total energy En.
SOLUTION The smallest value of En (i.e., the most negative) occurs when n = 6. Thus,
using Z = 1 for hydrogen, we find Z2
12
= – (13.6 eV ) 2 = –0.378 eV
n2
6
______________________________________________________________________________
En = – (13.6 eV ) 24. REASONING The orbital quantum number l can have any integer value from 0 up to
n – 1, so that it must be less than the principle quantum number n and it cannot be a negative
number. If for example, n = 4, l can have only the values 0, 1, 2, and 3.
The magnetic quantum number ml can only have integer values, including 0, from
– l to + l . Thus, it can have a negative value. For instance, if l = 2, ml can have the
values –2, –1, 0 ,+1, and +2.
SOLUTION Of the five states listed in the table, three are not possible. The ones that are
not possible, and the reasons they are not possible, are: 1560 THE NATURE OF THE ATOM State Reason (a) The quantum number l must be less than n (c) The quantum number ml must be less than or equal to l . (d) The quantum number l cannot be negative. ______________________________________________________________________________
25. REASONING
a. The ground state is the n = 1 state, the first excited state is the n = 2 state, and the second
excited state is the n = 3 state. The total energy (in eV) of a hydrogen atom in the n = 3
state is given by Equation 30.13.
b. According to quantum mechanics, the magnitude L of the angular momentum is given by
Equation 30.15 as L = l ( l + 1) ( h / 2π ) , where l is the orbital quantum number. The discussion in Section 30.5 indicates that the maximum value that l can have is one less than
the principal quantum number, so that l max = n – 1.
c. Equation 30.16 gives the zcomponent Lz of the angular momentum as Lz = ml ( h / 2π ) , where ml is the magnetic quantum number. According to the discussion in Section 30.5,
the maximum value that ml can attain is when it is equal to the orbital quantum number,
which is l max.
SOLUTION
a. The total energy of the hydrogen atom is given by Equation 30.13. Using n = 3, we have E3 = − (13.6 eV )(1)2
32 = −1.51 eV b. The maximum orbital quantum number is l max = n – 1 = 3 – 1 = 2. The maximum
angular momentum Lmax has a magnitude given by Equation 30.15:
Lmax = l max ( l max + 1) h
6.63 × 10−34 J ⋅ s
= 2 ( 2 + 1)
= 2.58 × 10−34 J ⋅ s
2π
2π c. The maximum value for the zcomponent Lz of the angular momentum is ( with ml = l max = 2 ) Lz = ml h
6.63 × 10−34 J ⋅ s
= ( 2...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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