CHAPTER 30 THE NATURE OF THE ATOM
CONCEPTUAL QUESTIONS
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1.
REASONING AND SOLUTION A tube is filled with atomic hydrogen at room temperature.
Electromagnetic radiation with a continuous spectrum of wavelengths, including those in the Lyman,
Balmer, and Paschen series, enters one end of the tube and leaves the other end. The exiting radiation
is found to contain absorption lines.
At room temperature, most of the atoms of atomic hydrogen contain electrons that are in the
ground state (n = 1) energy level.
Since the radiation contains a continuous spectrum of
wavelengths, it contains photons with a wide range of energies ( E = hf = hc / λ ). In particular, it will
contain photons with energies that are equal to the energy difference between the atomic states in the
Lyman series. When the radiation is incident on the atoms in the tube, these photons are absorbed by
the electrons. When a photon, whose energy is equal to the energy difference for a transition in the
Lyman series, is absorbed by a ground state electron, that electron will make a transition to a higher
energy level. Every photon of this energy will be absorbed by a ground state electron and cause a
transition. The wavelength of radiation that corresponds to that particular photon energy will,
therefore, be removed from the radiation. When the radiation is analyzed, the wavelengths that
correspond to transitions in the Lyman series will be absent. Since most of the atoms in the tube are
in the ground state (n = 1), the electron populations in the n = 2 and n = 3 states are extremely small.
Therefore, any absorption lines resulting from Balmer or Paschen transitions will be extremely weak.
When the radiation is analyzed, the only predominant absorption lines in the exiting radiation will
correspond to wavelengths in the Lyman series.
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2.
REASONING AND SOLUTION Refer to the situation described in Question 1. Suppose the
electrons in the atoms are mostly in excited states. Most of the electrons are in states with n > 2;
therefore, Balmer and Paschen series transitions will occur, and the absorption lines in the exiting
radiation will correspond to wavelengths in the Balmer and Paschen series. Since there are relatively
few electrons in the ground state, only a relatively few number of photons that correspond to
wavelengths in the Lyman series will be absorbed. Most of the "Lyman photons" will remain in the
radiation; therefore, the exiting radiation will not contain absorption lines that correspond to
wavelengths in the Lyman series. Although the absorption lines that correspond to transitions in the
Lyman series are not present, there will be more absorption lines in the exiting radiation compared to
the situation when the electrons are in the ground state, because absorption lines corresponding to
both the Balmer and Paschen series will be present.
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3.
REASONING AND SOLUTION
According to Equation 30.13, the energy En of the nth atomic
state in a Bohr atom is given by E n = – (13 .6 eV ) Z 2 / n 2 , where n = 1, 2, 3, . . ., and Z is equal to the
number of protons in the nucleus. The energy that must be absorbed
by the electron to cause an
upward transition from the initial state ni to the final
state nf is ∆ E fi = E f – E i . Using
(
)
Equation 30.13, we find that the required energy is ∆ E fi = – (13 .6 eV ) Z 2 1/ n 2f – 1 / n i2 , with ni, nf
= 1, 2, 3, . . . and n f > n i . The energy required to ionize the atom from when the outermost electron
Chapter 30 Conceptual Questions
175
is in the state ni can be found by letting the value of nf approach infinity. The ionization energy is
then
(
)
(
)
∆ E ∞ = – (13 .6 eV ) Z 2 1 / ∞ – 1 / n 2i = – (13 .6 eV ) Z 2 0 – 1 / n i2 = (13 .6 eV ) Z 2 / n i2
From this expression, we see that the ionization energy is inversely proportional to the square of the
principal quantum number ni of the initial state of the electron; thus, less energy is required to remove
the outermost electron in an atom when the atom is in an excited state. Therefore, when the atom is
in an excited state, it is more easily ionized than when it is in the ground state.
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4.
REASONING AND SOLUTION In the Bohr model for the hydrogen atom, the closer the electron
is to the nucleus, the smaller is the total energy of the atom. This is not true in the quantum
mechanical picture of the hydrogen atom.
In the Bohr model, the nth orbit is a circle of definite radius rn; every time that the position of the
electron in this orbit is measured, the electron is found exactly a distance rn from the nucleus. In
contrast, according to the quantum mechanical picture, when an atom is in the nth state, the position
of the electron is uncertain. Even though the atomic state is well defined by the principal quantum
number n, the location of the electron is not definite. When the atom is in the nth state, the electron
can sometimes be found close to the nucleus, while, at other times, it can be found far from the
nucleus, or at some intermediate position. In the absence of any external magnetic fields, the energy
of the state is determined by the value of n, and for a given n, the position of the electron is uncertain.
Therefore, it is not correct to say that in the quantum mechanical picture of the hydrogen atom, the
closer the electron is to the nucleus, the smaller is the total energy of the atom.
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5.
REASONING AND SOLUTION
a. Consider two different hydrogen atoms. The electron in each atom is in an excited state. The
Bohr model uses the same quantum number n to specify both the energy and the orbital angular
momentum. According to the Bohr model, if an atom is in its nth state, the corresponding energy is
E n = − (13 .6 eV ) Z 2 / n 2 , where n = 1, 2, 3, . . ., and Z is equal to the number of protons in the
nucleus (Equation 30.13). The integer n that specifies the energy also specifies the orbital angular
momentum according to Ln = nh /(2 π ) , where h is Planck's constant (Equation 30.8). Therefore,
according to the Bohr model, it is not possible for the electrons to have different energies, but the
same orbital angular momentum.
b. The quantum mechanical model uses two different quantum numbers to specify the energy and the
orbital angular momentum of an atomic electron. In the absence of magnetic fields, the energy is
determined by the principal quantum number n, while the orbital angular momentum is determined by
the orbital quantum number l. According to quantum mechanics, if an atom is in its nth state with
orbital quantum number l, the energy is identical to that of the Bohr model, and the magnitude of the
orbital angular momentum is [h / ( 2 π )] l ( l + 1 ) , where l = 0, 1, 2, . . . , (n – 1). Since l = 0, 1, 2, . .
. , (n – 1), different values of l are compatible with the same value of n. For example, when n = 3,
the possible values of l are 0, 1, 2, and when n = 4, the possible values of l are 0, 1, 2, 3. Thus, the
176
THE NATURE OF THE ATOM
electron in one of the atoms could have n = 3, l = 2, while the electron in the other atom could have n
= 4, l = 2. Therefore, according to quantum mechanics, it is possible for the electrons to have
different energies but have the same orbital angular momentum.
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6.
REASONING AND SOLUTION In the quantum mechanical picture of the hydrogen atom, the
orbital angular momentum of the electron may be zero in any of the possible energy states. This is
true because, if an atom is in its nth state with orbital quantum number l, the magnitude of the orbital
angular momentum is given by [h / ( 2 π )] l ( l + 1 ) , where l = 0, 1, 2, . . . , (n – 1); therefore, for any
value of n, the value of l may be zero. However, when the hydrogenic electron is in the ground state,
n = 1, and the only possible value for the orbital quantum number is l = 0. Therefore, when the
electron is in the ground state, its orbital angular momentum must be zero.
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7.
REASONING AND SOLUTION
a. According to the Pauli exclusion principle, no two electrons in an atom can have the same set of
values for the four quantum numbers n, l, ml, and ms. For a given value of the orbital quantum
number l, the magnetic quantum number ml, can have 2l + 1 possible values. Each of these values
can be combined with two possible values for the spin quantum number ms. Therefore, the maximum
number of electrons that the lth subshell can hold is 2(2l + 1). All of the electrons in a 5g subshell
have n = 5, and l = 4. Thus, the maximum number of electrons that can be contained in a 5g subshell
is 2(2l + 1) = 2[2(4) + 1] = 18. Therefore, a 5g subshell can contain at most 18 electrons and cannot
contain 22 electrons.
b. Since 17 is less than 18, a 5g subshell could contain 17 electrons.
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8.
REASONING AND SOLUTION An electronic configuration for manganese (Z = 25) is written as
2
2
6
2
6
4
2
1
According to Figure 30.17, the ground state electronic
1s 2s 2p 3s 3p 3d 4s 4p .
configuration for manganese is written as 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 4s 2 . Clearly, the electronic
configuration given in the question corresponds to an excited state in which the last electron in the 3d
subshell has been excited to the 4p subshell.
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9.
REASONING AND SOLUTION Characteristic X-rays are produced when an electron with enough
energy strikes the target atom and knocks one of the K-shell ( n = 1 ) electrons entirely out of the atom.
An electron in one of the outer shells ( n ≥ 2 ) then drops into the K-shell and emits an X-ray photon in
the process.
The ground state of hydrogen contains one electron in the n = 1 state, whereas, the ground state of
helium contains two electrons in the n = 1 state. In both atoms, the ground state configuration
contains only K-shell electrons. Even if one of the K-shell electrons were knocked out by an
impinging electron, there are no electrons in higher levels to fall into the K-shell vacancy. Therefore,
we would not expect hydrogen and helium atoms in their ground state to emit characteristic X-rays.
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Chapter 30 Conceptual Questions
177
10. REASONING AND SOLUTION The characteristic lines in the X-ray spectrum depend on the
target material. They occur at the wavelength value that corresponds to the wavelength of the X-ray
photon that was emitted when an electron from the outer shell of the atom in the target material drops
down to the K-shell. Since all identical atoms of the same element contain the same energy levels,
the Kα and Kβ lines will occur at the same wavelength values for all X-ray targets made of those
atoms, regardless of the voltage applied across the X-ray tube.
The cutoff wavelength λ0, however, depends on the voltage V across the X-ray tube. According to
Equation 30.17, λ 0 = ( hc ) /( eV ) ; the cutoff wavelength is inversely proportional to the voltage
across the tube. Therefore, when a smaller voltage is used to operate the tube, the cutoff wavelength
increases.
We can conclude, therefore, that when the tube is operated at a smaller voltage, the characteristic
lines should occur at the same wavelengths as they did at a higher voltage, and the cutoff wavelength
should increase (i.e., shift to the right on the wavelength axis). The drawing in the text, shows the Xray spectra produced by an X-ray tube when the tube is operated at two different voltages. The
features of the drawing are consistent with the conclusions reached above.
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11. REASONING AND SOLUTION
Bremsstrahlung X-rays are produced when the electrons
decelerate upon hitting the target. In the production of X-rays, it is possible to create Bremsstrahlung
X-rays without producing the characteristic X-rays. The wavelengths of the characteristic lines
depend on the target material; the cutoff wavelength, however, is independent of the target material
and depends only on the energy of the impinging electrons. In order to produce Bremsstrahlung Xrays without producing the characteristic X-rays, the potential difference V applied to the tube must
be chosen so that the cutoff wavelength is greater than the wavelengths of the characteristic X-rays
for the target material. Therefore, using Equation 30.17, the value of V must be chosen so that
V = ( hc ) /( e λ 0 ) , where λ 0 > λ K .
α
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12. REASONING AND SOLUTION
The short side of X-ray spectra ends abruptly at a cutoff
wavelength λ0. This cutoff wavelength depends only on the energy of the impinging electrons.
Regardless of the composition of the target, the incident electron cannot give up more than all of its
kinetic energy when it is decelerated by the target. The kinetic energy that the electron has acquired
upon reaching the target is given by eV (see Section 19.2), where V is the potential difference that is
applied to the X-ray tube. Therefore, the cutoff wavelength depends on the value of V, and not on the
composition of the target material used in the X-ray tube.
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13. REASONING AND SOLUTION A laser beam focused to a small spot can cut through a piece of
metal. Light rays leaving the laser are nearly parallel, so the light can be focused to a very small spot.
The intensity is the electromagnetic energy per unit time per unit area (see Equation 24.4). Therefore,
when the beam is focused to a spot with a very small area, the intensity delivered by the beam is very
large. For example, a pulsed ruby laser with a peak power of 108 W can be focused to a peak
intensity of 1020 W/cm2. This intensity is large enough to ionize air or burn a hole in a piece of
metal.
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178
THE NATURE OF THE ATOM
14. REASONING AND SOLUTION The energy of a photon of light of frequency f is given by
Equation 29.2: E = hf . Since c = λ f for light, the energy of a photon can be written in terms of the
wavelength λ as E = hc / λ . According to Table 26.2, the wavelength of green light is smaller than
the wavelength of red light; therefore, the laser that produces green light emits photons that are more
energetic than those emitted by the helium/neon laser.
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