Transcript Document
Example 7.1
Calculate the frequency of an X ray that has a wavelength of 8.21 nm.
Strategy
When using Equation (7.4), the value of
c
will always be 3.00 x 10 8 m s –1 . We must solve Equation (7.4) for the unknown variable, either
v
or and substitute a known value for the other variable, expressed in the appropriate unit. Here,
v
is the unknown; is the known value, expressed in meters.
Solution
First, we solve Equation (7.4) for
v
.
Next, we convert the wavelength from nanometers to meters.
Then, we can substitute the known values of
c
and .
Assessment
Besides checking for correct units when applying Equation (7.4), expect
v
generally to be a large number, often a large positive power of ten, and expect (in meters) generally to be a small number, often a large negative power of ten.
Example 7.1 continued Exercise 7.1A
Calculate the frequency, in hertz, of a microwave that has a wavelength of 1.07 mm.
Exercise 7.1B
Calculate the wavelength, in nanometers, of infrared radiation that has a frequency of 9.76 x 10 13 Hz.
Example 7.2
Which light has the higher frequency: the bright red brake light of an automobile or the faint green light of a distant traffic signal?
Analysis and Conclusions
The difference in brightness is immaterial because, as we have noted, the amplitude of a light wave has no bearing on the frequency. From Figure 7.10, we see that there is a relationship between the color of light and its frequency and wavelength. The order of increasing frequency in the visible spectrum proceeds from red to violet: Red < orange < yellow < green < blue < violet The green light has a higher frequency than the red light.
Exercise 7.2A
Which source produces electromagnetic radiation of the longer wavelength: a microwave oven or the fluorescent screen of a color television set?
Exercise 7.2B
Which of the following has the highest frequency: (a) green light, (b) radiation with wavelength 4610 Å, (c) 91.9 MHz radiation, (d) 622-nm radiation?
Example 7.3
Calculate the energy, in joules, of a photon of violet light that has a frequency of 6.15 x 10 14 s –1 .
Strategy
This problem involves a direct application of Planck’s Equation (7.5), where both Planck’s constant,
h,
and the frequency,
v
, are known.
Solution Exercise 7.3A
Calculate the energy, in joules per photon, of microwave radiation that has a frequency of 2.89 x 10 10 s –1 .
Exercise 7.3B
Calculate the energy, in joules per photon, of ultraviolet light with a wavelength of 235 nm.
Example 7.4
A laser produces red light of wavelength 632.8 nm. Calculate the energy, in kilojoules, of 1 mol of photons of this red light.
Strategy
We know that Equation (7.5) yields the energy of one photon, but for 1 mol photons we must use Equation (7.6). However, we first need to find the frequency of the red light; and to obtain this, we need Equation (7.4), remembering also that wavelength must be expressed in meters. At this point, we will have all the necessary data to apply Equation (7.6).
Solution
First, we convert a 632.8-nm wavelength to meters. Next, we use this wavelength in Equation (7.4) to obtain the frequency of the red light.
Finally, we substitute Avogadro’s number, Planck’s constant, and the light frequency into Equation (7.6).
Example 7.4 continued Assessment
In this problem it is very important to check the signs of the powers of ten to be certain that they conform to the quantities they represent. That is, the wavelength of red light in meters is a very small number; the frequency is a large number; Avogadro’s number is very large; and Planck’s constant is very small. Using powers of ten with the wrong sign can cause errors of many orders of magnitude.
Exercise 7.4A
The lower wavelength limit of visible light is about 400 nm. What is the energy of this radiation, expressed in kilojoules per mole of photons?
Exercise 7.4B
Use Figure 7.10 to help you determine in which region of the electromagnetic spectrum you would find radiation having an energy of 100 kJ/mol.
Example 7.5
Calculate the energy of an electron in the second energy level of a hydrogen atom.
Solution
We can use Equation (7.7) for the hydrogen atom, with
B n
= 2: = 2.179 x 10 –18 J and
Exercise 7.5A
Calculate the energy of an electron in the energy level
n
= 6 of a hydrogen atom.
Exercise 7.5B
Is there a hydrogen-atom energy level at –2.179 x 10 –19 J? Explain.
Example 7.6
Calculate the energy change, in joules, that occurs when an electron falls from the
n
i to the
n
f = 3 energy level in a hydrogen atom.
= 5
Strategy
We find the initial and final energy levels of the electron in the statement of the problem. To determine the energy change during an electron transition, we then substitute the numbers of the levels (
n
i = 5 and
n
f = 3) and the value of the constant
B
(2.179 x 10 –18 J) into Equation (7.8).
Solution Assessment
The minus sign indicates that the atom has given up energy in the form of a photon of light, corresponding to the drop from the higher to lower energy level. When using Equation (7.8), always compare your result with your intuitive expectation.
Exercise 7.6A
Calculate the energy change that occurs when an electron is raised from the
n
i level to the
n
f = 4 level of a hydrogen atom.
= 2 energy
Exercise 7.6B
Can there be an energy difference of 4.269 x hydrogen atom? Explain.
10 –20 J between energy levels for the
Example 7.7
Calculate the frequency of the radiation released by the transition of an electron in a hydrogen atom from the
n
= 5 level to the
n
= 3 level, the transition we looked at in Example 7.6.
Solution
From Example 7.6, we know that ∆
E
level the value of
E
level = 1.550 x 10 –19 J. Using this quantity as in Equation (7.9), we can solve the equation for
v
:
Assessment
In Example 7.6, we obtained ∆ sign and used only the magnitude: –1.550 surroundings (positive).
E
level = –1.550 x 10 x 10 –19 –19 J. The negative sign indicated that energy was emitted. Here in Example 7.7, we dropped the negative J. We did this so that
v
would be a positive quantity. We can accomplish the same objective by considering that the excited atom loses energy (negative) and the emitted photon carries it to the
Exercise 7.7A
Calculate the frequency of the radiation emitted during the transition of an electron from the
n
= 4 energy level in a hydrogen atom to the
n
= 1 level.
Exercise 7.7B
Calculate the wavelength, in nanometers, of the radiation emitted as the electron in a hydrogen atom moves from the
n
= 5 level to the
n
= 2 level. In what region of the electromagnetic spectrum is the spectral line produced by this radiation?
Example 7.8 Conceptual Example
Without doing detailed calculations
spectrum.
, determine which of the four electron transitions shown in Figure 7.19 produces the shortest-wavelength line in the hydrogen emission
Example 7.8 Conceptual Example continued Analysis and Conclusions
First, we should recognize that transition
(a),
even though it involves the greatest span of energy levels, requires energy change, the
greater absorption,
not emission. We can therefore eliminate (a). The other three transitions do involve light emission. Let’s compare the energy changes in these transitions, recognizing that the larger the energy the frequency and the
shorter
the wavelength of the spectral line produced. Transitions
(b)
and
(d)
both terminate at the same level,
n
= 2, but the drop in energy in
(b)
is greater than in
(d).
level as transition
(b),
Transition
(c)
begins at the same but terminates at a higher energy level,
n
= 3. The energy change for transition (c) is therefore less than that for transition
(b).
Thus, the shortest-wavelength spectral line of the three is that produced by transition
(b).
Exercise 7.8A
Without doing detailed calculations,
(a)
from
n
= 1 to
n
= 2,
(b)
from
n
determine which of the following electron transitions in a hydrogen atom requires absorption of the greatest amount of energy: = 3 to
n
= 8,
(c)
from
n
= 4 to
n
= 1,
(d)
from
n
= 2 to
n
= 3.
Exercise 7.8B
Can the electron in a hydrogen atom drop from the level
n
4.90 x 10 –20 J of energy? Explain. = 4 with the emission of
Example 7.9
Calculate the wavelength, in meters and nanometers, of an electron moving at a speed of 2.74 x 10 6 m/s. The mass of an electron is 9.11 x 10 –31 kg, and 1 J = 1 kg m 2 s –2 .
Strategy
To use de Broglie’s equation (7.10), we must first substitute the units kg m 2 the energy unit J. This changes the units of Planck’s constant,
h,
s –2 for from J s to kg m 2 s – 2 s. Then we must make sure that the particle mass is in kilograms and the particle velocity is in meters per second. These required units are used in the statement of this particular problem.
Solution
We can substitute the given data directly into the de Broglie equation: The wavelength of the electron is 2.65 x 10 –10 m or 0.265 nm.
Assessment
As we have implied, the cancellation of units leading to a wavelength in meters will not occur unless we have converted the other data in Equation (7.10) to the proper units. A further check on the reasonableness of the answer is that the wavelength must be short, for example, in the nanometer range.
Example 7.9 continued Exercise 7.9A
Calculate the wavelength, in nanometers, of a proton moving at a speed of 3.79 x 10 3 m/s. The mass of a proton is 1.67 x 10 –27 kg.
Exercise 7.9B
Use data from Example 7.9 and Exercise 7.9A to determine the speed a proton must have if its wavelength is to be equal to that of an electron moving at 2.74 x 10 6 m/s.
Example 7.10
Considering the limitations on values for the various quantum numbers, state whether an electron can be described by each of the following sets. If a set is not possible, state why not.
(a)
n
= 2,
l
= 1,
m l
(b)
n
= 1,
l
= 1,
m l
= –1 = +1
(c) (d)
n
= 7,
l n
= 3,
l
= 3,
m l
= 1,
m l
= +3 = –3
Solution
(a) (b) (c) (d)
All the quantum numbers are allowed values.
Not possible. The value of
l
must be less than the value of
n
.
All the quantum numbers are allowed values.
Not possible. The value of
m l
or +1).
must be in the range –
l
to +
l
(in this case, –1, 0,
Exercise 7.10A
Considering the limitations on values for the various quantum numbers, state whether an electron can be described by each of the following sets. If a set is not possible, state why not.
(a)
n
= 2,
l
= 1,
m l
(b)
n
= 3,
l
= 2,
m l
= –2 = +2
(c)
n
= 4,
l
(d)
n
= 5,
l
= 3,
m l
= 2,
m l
= +3 = +3
Exercise 7.10B
Replace the question marks by suitable responses in the following quantum number assignments.
(a)
n
= 3,
l
= 1,
m l
= ?
(b)
n
= 4,
l
= ?,
m l
= –2
(c)
n
= ?,
l
= 3,
m l
= ?
Example 7.11
Consider the relationship among quantum numbers and orbitals, subshells, and principal shells to answer the following.
(a)
How many orbitals are there in the 4
d
subshell?
(b)
What is the first principal shell in which
f
orbitals can be found? (
c)
Can an atom have a 2
d
subshell?
(d)
Can a hydrogen atom have a 3
p
subshell?
Solution
(a) (b) (c) (d)
The
d
possible values of the 4
d
subshell corresponds to a value of subshell.
m l l
= 2. With
l
= 2, there are five : –2, –1, 0, +1, and +2. Thus, there are five
d
orbitals in The
f
orbital corresponds to a value of
l
is
n
– 1, the allowable values of
n
= 3. Because the maximum value of are 4, 5, 6, and so on. The first shell to
l
contain
f
orbitals is therefore the fourth principal shell (
n
No. For a
d
subshell,
l
= 2. The maximum value of
l
is
n
cannot be 2. There cannot be a 2
d
subshell.
= 4).
– 1, and so if
n
= 2,
l
Yes. Although the electron in a hydrogen atom is usually in the 1
s
orbital, it can be excited to one of the higher energy states, such as one of the 3
p
orbitals.
Exercise 7.11A
Consider the relationship among quantum numbers and orbitals, subshells, and principal shells to answer the following.
(a)
How many orbitals are there in the 5
p
subshell?
(b)
What subshells of the hydrogen atom consist of a total of seven orbitals?
Exercise 7.11B
What is the total number of orbitals in the principal shell for which quantum number
n
.
n
= 4? Derive an equation that relates the total number of orbitals in a principal shell to the principal
Cumulative Example
Which will produce more energy per gram of hydrogen: H atoms undergoing an electronic transition from the
n
= 4 level to the
n
= 1 level or hydrogen gas burned in the reaction 2 H 2 (g) + O 2 (g) 2 H 2 O(l)?
Strategy
We can use Equation (7.8) to find the photon energy emitted by one hydrogen atom as its electron undergoes a transition from H 2
n
= 4 to
n
= 1. We can multiply the energy of one photon by Avogadro’s number to determine the energy emitted by 1 mol of hydrogen atoms. Then we can use the molar mass to find the energy produced per gram of hydrogen. In a separate calculation, we can determine the enthalpy change first for the combustion reaction as written and then per gram of (g) burned. Our final step is to compare the two results.
Solution
Energy emitted as electromagnetic radiation:
We first apply Equation (7.8) to the transition from
n
= 4 to
n
= 1.
Cumulative Example continued Solution continued
The calculated energy is for one atom. We multiply by Avogadro’s number to obtain the energy emitted per mole of H, and then convert from moles of H to grams of H.
Enthalpy of combustion of H
formation of H 2 O(l).
2 (g):
To obtain this enthalpy of combustion, we can use Equation (6.21), but because the enthalpies of formation of H 2 (g) and O 2 (g) are zero, we note that the enthalpy of combustion is simply twice the enthalpy of Now we convert this enthalpy of combustion to a per-gram H 2 (g) basis.
Comparison of results:
The energy emitted per gram of hydrogen as electromagnetic radiation by H atoms (1221 kJ) is over eight times as great as that released in the combustion of H 2 (g) (141.8 kJ).
Cumulative Example continued Assessment
Figure 7.18 shows that the emitted light falls in the ultraviolet region of the spectrum. Example 7.4 provides information to help assess our results. In that example, light of wavelength 632.8 nm was found to have an energy of 189.1 kJ/mol photons. Energy is inversely proportional to wavelength ( wavelength associated with the
n
= 4
E n
=
hv
=
hc
/ ), and the = 1 transition, 97.3 nm, is less than one sixth of 632.8 nm. Thus, the energy we calculate for this transition should be more than six times 189 kJ/mol, or roughly 1200 kJ/mol. Hydrogen has a molar mass very nearly 1 g H/mol H, which corresponds to 1200 kJ/g H.