The Quantum Theory of the Submicroscopic World

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chapter 1

The quantum theory of the submicroscopic world

1.1(a)

(b)

1.2(a)

Strategy: We are given the wavelength of an electromagnetic wave and asked to calculate the frequency. Rearranging Equation (1.2) of the text and replacing u with c (the speed of light) gives:

Solution: Because the speed of light is given in meters per second, it is convenient to first convert wavelength to units of meters. Recall that 1 nm  1  109 m. We write:

Substituting in the wavelength and the speed of light (), the frequency is:

Check: The answer shows that 6.58  1014 waves pass a fixed point every second. This very high frequency is in accordance with the very high speed of light.

(b)

Strategy: We are given the frequency of an electromagnetic wave and asked to calculate the wavelength. Rearranging Equation (1.2) of the text and replacing u with c (the speed of light) gives:

Solution: Substituting in the frequency and the speed of light () into the above equation, the wavelength is:

The problem asks for the wavelength in units of nanometers. Recall that 1 nm  1  109 m.

1.3

This radiation falls in the microwave region of the spectrum. (See Figure 1.6 of the text.)

1.4The wavelength is:

1.5

1.6(a)

Strategy: We are given the frequency of an electromagnetic wave and asked to calculate the wavelength. Rearranging Equation (1.2) of the text and replacing u with c (the speed of light) gives:

Solution: Substituting in the frequency and the speed of light (3.00  108 m s-1) into the above equation, the wavelength is:

Check: The wavelength of 400 nm calculated is in the blue region of the visible spectrum as expected.

(b)

Strategy: We are given the frequency of an electromagnetic wave and asked to calculate its energy. Equation (1.3) of the text relates the energy and frequency of an electromagnetic wave.

Eh

Solution: Substituting in the frequency and Planck's constant (6.63  1034 J s) into the above equation, the energy of a single photon associated with this frequency is:

Check: We expect the energy of a single photon to be a very small energy as calculated above, 5.0  1019 J.

1.7(a)

The radiation does not fall in the visible region; it is radio radiation. (See Figure 1.6 of the text.)

(b)E  h  (6.63  1034 J s)(6.0  104 s-1)  4.0  1029 J

(c)Converting to J mol-1:

1.8The energy given in this problem is for 1 mole of photons. To apply Eh, we must divide the energy by Avogadro’s number. The energy of one photon is:

The wavelength of this photon can be found using the relationship,

The radiation is in the ultraviolet region (see Figure 1.6 of the text).

1.9

1.10(a)

(b)Checking Figure 1.6 of the text, you should find that the visible region of the spectrum runs from 400 to 700 nm. 370 nm is in the ultraviolet region of the spectrum.

(c)E  h. Substitute the frequency () into this equation to solve for the energy of one quantum associated with this frequency.

1.11First, we need to calculate the energy of one 600 nm photon. Then, we can determine how many photons are needed to provide 4.0  1017 J of energy.

The energy of one 600 nm photon is:

The number of photons needed to produce 4.0  1017 J of energy is:

1.12The heat needed to raise the temperature of 150 mL of water from 20C to 100C is:

heat  (150 ml)(4.184 J ml-1C-1)(100  20)C  5.0  104 J

The microwave will need to supply more energy than this because only 92.0% of microwave energy is converted to thermal energy of water. The energy that needs to be supplied by the microwave is:

The energy supplied by one photon with a wavelength of 1.22  108 nm (0.122 m) is:

The number of photons needed to supply 5.4  104 J of energy is:

1.13This problem must be worked to four significant figure accuracy. We use 6.6256  1034 Js for Planck’s constant and for the speed of light. First calculate the energy of each of the photons.

For one photon the energy difference is:

E  (3.372  1019 J)  (3.369  1019 J)  3  1022 J

For one mole of photons the energy difference is:

1.14The radiation emitted by the star is measured as a function of wavelength (). The wavelength corresponding to the maximum intensity is then plugged in to Wien’s law (Tmax = 1.44 10-2 K m) to determine the surface temperature.

1.15Wien’s law:Tmax = 1.44 10-2 Km.

Solving for temperature (T) yields:

100nm: ultraviolet

300nm: ultraviolet

500nm: visible (green)

800nm: infrared

Most night vision goggles work by converting low energy infrared radiation (which our eyes can not detect) to higher energy visible radiation.

1.16Because the same number of photons are being delivered by both lasers and because both lasers produce photons with enough energy to eject electrons from the metal, each laser will eject the same number of electrons. The electrons ejected by the blue laser will have higher kinetic energy because the photons from the blue laser have higher energy.

1.17In the photoelectric effect, light of sufficient energy shining on a metal surface causes electrons to be ejected (photoelectrons). Since the electrons are charged particles, the metal surface becomes positively charged as more electrons are lost. After a long enough period of time, the positive surface charge becomes large enough to increase the amount of energy needed to eject electrons from the metal.

1.18(a)The maximum wavelength is obtained by solving Equation 1.4 when the kinetic energy of the ejected electrons is equal to zero. In this case, Equation 1.4, together with the relation , gives

(b)

Please remember to carry extra significant figures during the calculation and then round at the very end to the appropriate number of significant figures.

1.19 The minimum (threshold) frequency is obtained by solving Equation 1.4 when the kinetic energy of the ejected

electrons is equal to zero. In this case, Equation 1.4 gives

1.20The emitted light could be analyzed by passing it through a prism.

1.21Light emitted by fluorescent materials always has lower energy than the light striking the fluorescent substance. Absorption of visible light could not give rise to emitted ultraviolet light because the latter has higher energy.

The reverse process, ultraviolet light producing visible light by fluorescence, is very common. Certain brands of laundry detergents contain materials called “optical brighteners” which, for example, can make a white shirt look much whiter and brighter than a similar shirt washed in ordinary detergent.

1.22Excited atoms of the chemical elements emit the same characteristic frequencies or lines in a terrestrial laboratory, in the sun, or in a star many light-years distant from earth.

1.23(a)The energy difference between states E1 and E4 is:

E4E1  (1.0  1019 J)  (15  1019 J)  14  1019 J

(b)The energy difference between the states E2 and E3 is:

E3E2  (5.0  1019 J)  (10.0  1019 J)  5  1019 J

(c)The energy difference between the states E1 and E3 is:

E1E3  (15  1019 J)  (5.0  1019 J)  10  1019 J

Ignoring the negative sign of E, the wavelength is found as in part (a).

1.24We use more accurate values of h and c for this problem.

1.25In this problem ni 3 and nf 5.

The sign of E means that this is energy associated with an absorption process.

Is the sign of the energy change consistent with the sign conventions for exo- and endothermic processes?

1.26Strategy: We are given the initial and final states in the emission process. We can calculate the energy of the emitted photon using Equation (1.5) of the text. Then, from this energy, we can solve for the frequency of the photon, and from the frequency we can solve for the wavelength. The value of Rydberg's constant is
2.18  1018 J.

Solution: From Equation (1.5) we write:

E  4.09  1019 J

The negative sign for E indicates that this is energy associated with an emission process. To calculate the frequency, we will omit the minus sign for E because the frequency of the photon must be positive. We know that

E  h

Rearranging this equation and substituting in the known values,

We also know that. Substituting the frequency calculated above into this equation gives:

Check: This wavelength is in the visible region of the electromagnetic region (see Figure 1.6 of the text). This is consistent with the transition from ni 4 to nf 2 gives rise to a spectral line in the Balmer series (see Figure 1.15 of the text).

1.27The Balmer series corresponds to transitions to the n 2 level.

For He:

For the transition, n 6  2

For the transition, n 6  2, E1.94  1018 J 103 nm

For the transition, n 5  2, E1.83  1018 J 109 nm

For the transition, n 4  2, E1.64  1018 J 121 nm

For the transition, n 3  2, E1.21  1018 J 164 nm

For H, the calculations are identical to those above, except the Rydberg constant for H is 2.18  1018 J.

For the transition, n 6  2, E4.84  1019 J 411 nm

For the transition, n 5  2, E4.58  1019 J 434 nm

For the transition, n 4  2, E4.09  1019 J 487 nm

For the transition, n 3  2, E3.03  1019 J 657 nm

All the Balmer transitions for He are in the ultraviolet region; whereas, the transitions for H are all in the visible region. Note the negative sign for energy indicating that a photon has been emitted.

1.28Equation 1.14:

for n = 2:

for n = 3:

1.29ni  236, nf  235

This wavelength is in the microwave region. (See Figure 1.6 of the text.)

1.30

1.31First convert the parameters to the appropriate units and then use the following equation:

(a)

(b)

1.32

1.33Strategy: We are given the mass and the speed of the proton and asked to calculate the wavelength. We need the de Broglie equation, which is Equation 1.20 of the text. Note that because the units of Planck's constant are J s, m must be in kg and u must be in m s1 (1 J  1 kg m2 s2).

Solution: Using Equation 1.20 we write:

The problem asks to express the wavelength in nanometers.

1.34Converting the velocity to units of ms1:

1.35First, we convert mph to m s1.

1.36From the Heisenberg uncertainty principle (Equation 1.22) we have

For the minimum uncertainty we can change the greater than sign to an equal sign.

Because p = m u, the uncertainty in the velocity is given by

The minimum uncertainty in the velocity is 1 106 m s–1.

1.37Below are the probability distributions for the first three energy levels for a one-dimensional particle in a box, as shown in Figure 1.24. The average value for x for all of the states for a one-dimensional particle in a box is L/2 (the middle of the box). This is true even for those states (like the second state) that have no probability of being exactly in the middle of the box.

1.38The difference between the energy levels n and (n + 1) for a one-dimensional particle in a box is given by Equation 1.31:

.

All we have to do is to plug the numbers into the equation remembering that 1Å=10-10m.

Energy is equal to Planck’s constant times frequency. We can rearrange this equation to solve for frequency.

1.39From Equation 1.15 we can derive an equation for the change in energy going from the n=1 to n=2 states for a hydrogen atom:

.

From Equation 1.31 we can derive an equation for the change in energy going from the n=1 to n=2 states for a particle in a box:

.

We can combine the two equations and solve for L.

The radius of the ground state of the hydrogen atom (the Bohr radius) is 5.29  10-11 m (see Example 1.4). This leads to a diameter of 1.06  10-10 m, which is only a third the size of L that we calculated. So, it seems like a crude estimate.

1.40

The average value for x will be greater than L/2. The larger the electric field, the larger the average value of x.

1.41(a)The number given in front of the designation of the subshell is the principal quantum number, so in this case n3. For s orbitals, l 0. ml can have integer values from l to l, therefore, ml 0. The electron spin quantum number, ms, can be either 1/2 or −1/2.

Following the same reasoning as part (a)

(b)4p: n 4; l 1; ml −1, 0, or 1; ms1/2, −1/2

(c)3d: n 3; l 2; ml −2, −1, 0, 1, or 2; ms1/2, −1/2

An orbital in a subshell can have any of the allowed values of the magnetic quantum number for that subshell. All the orbitals in a subshell have exactly the same energy.

1.42(a)2p: n 2, l 1, ml 1, 0, or 1

(b)6s: n 6, l 0, ml 0 (only allowed value)

(c)5d: n 5, l 2, ml 2, 1, 0, 1, or 2

An orbital in a subshell can have any of the allowed values of the magnetic quantum number for that subshell. All the orbitals in a subshell have exactly the same energy.

1.43The allowed values of l are 0, 1, 2, 3, and 4. These correspond to the 5s, 5p, 5d, 5f, and 5g subshells. These subshells each have one, three, five, seven, and nine orbitals, respectively.

1.44For n 6, the allowed values of l are 0, 1, 2, 3, 4, and 5 [l 0 to (n 1), integer values]. These l values correspond to the 6s, 6p, 6d, 6f, 6g, and 6h subshells. These subshells each have 1, 3, 5, 7, 9, and 11 orbitals, respectively (number of orbitals  2l 1).

1.45There can be a maximum of two electrons occupying one orbital.

(a)two(b)six(c)ten(d)fourteen

What rule of nature demands a maximum of two electrons per orbital? Do they have the same energy? How are they different? Would five 4d orbitals hold as many electrons as five 3d orbitals? In other words, does the principal quantum number n affect the number of electrons in a given subshell?

1.46A 2s orbital is larger than a 1s orbital. Both have the same spherical shape. The 1s orbital is lower in energy than the 2s. The 1s orbital does not have a node while the 2s orbital does.

1.47The two orbitals are very similar in shape, though the 3py is larger and has a higher energy. The 3pyorbital will have 1 radial node, in contrast to the 2pxorbital, which has no radial nodes. The two orbitals also differ in their orientation: 3pyis oriented parallel to the y-axis and 2pxis oriented parallel to the x-axis. Can you assign a specific value of the magnetic quantum number to these orbitals? What are the allowed values of the magnetic quantum number for the 2p subshell?

1.48We can modify Equation 1.15 to determine the ionization energies of single electron atoms. Ionization is the process of removing an electron (exciting it to a infinitely high energy state).

He+ has two protons in the nucleus and hence a +2 nuclear charge, making Z=2.

We need to convert our answer to kJ mol–1.

Li2+ has three protons in the nucleus and hence a +3 nuclear charge, making Z = 3.

We need to convert our answer to kJ mol–1.

1.49Equation 1.15 yields the energy of the photon absorbed for different transitions. For an emission, Equation 1.15 can be written:

Energy equals Planck’s constant times the speed of light divided by the wavelength: .

We can use this relationship to modify the above equation.

We can now solve for one over ninitial.

Hence, ninitial is 5.

1.50(a)False. n 2 is the first excited state.

(b)False. In the n 4 state, the electron is (on average) further from the nucleus and hence easier to remove.

(c)True.

(d)False. The n 4 to n 1 transition is a higher energy transition, which corresponds to a shorterwavelength.

(e)True.

1.51The energy given in the problem is the energy of 1 mole of gamma rays. We need to convert this to the energy of one gamma ray, and then we can calculate the wavelength and frequency of this gamma ray.

Now, we can calculate the wavelength and frequency from this energy.

and

E  h

1.52We first calculate the wavelength, then we find the color using Figure 1.6 of the text.

1.53(a)First, we can calculate the energy of a single photon with a wavelength of 633 nm.

The number of photons produced in a 0.376 J pulse is:

(b)Since a 1 W  1 Js-1, the power delivered per a 1.00  109 s pulse is:

Compare this with the power delivered by a 100-W light bulb!

1.54The energy required to heat the water is:

mCsT  (368 g)(4.184 J g-1C-1)(5.00C)  7.70  103 J,

where m is the mass, Cs the specific heat of water, and T is the change in temperature.

The energy of a photon with a wavelength of 1.06  104 nm can be calculated as follows:

The number of photons required is:

1.55First, let’s find the energy needed to photodissociate one water molecule.

The maximum wavelength of a photon that would provide the above energy is:

This wavelength is in the visible region of the electromagnetic spectrum. Since water is continuously being struck by visible radiation without decomposition, it seems unlikely that photodissociation of water by this method is feasible.

1.56For the Lyman series nf = 1. The longest wavelength emission corresponds to the smallest change in energy (ni 2). By multiplying both sides of Equation 1.5 by Planck’s constant the following equation is obtained:

For the Balmer series nf = 2, the shortest wavelength emission corresponds to the largest change in energy (ni).

The longest wavelength emission for the Lyman series (121 nm) is shorter than the shortest wavelength emission for the Balmer series (365 nm). Hence, the two series do not overlap.

1.57Since 1 W  1 J s-1, the energy output of the 75-W light bulb in 1 second is 75 J. The actual energy converted to visible light is 15 percent of this value (0.15 75 = 11 J).

First, we need to calculate the energy of one 550 nm photon. Then, we can determine how many photons are needed to provide 11 J of energy.

The energy of one 550 nm photon is:

The number of photons needed to produce 11 J of energy is:

1.58The energy needed per photon for the process is:

Any wavelength shorter than 483 nm will also promote this reaction. Once a person goes indoors, the reverse reaction Ag  Cl  AgCl takes place.

1.59Starting with Equation 1.4,

,

we can set the kinetic energy equal to eV,

,

divide both sides by h to yield:

.

To use the equation above to determine h and  we plot as a function of V. The slope will be equal to e/h and the y-intercept will be equal to  /h. Before we can plot, we have to calculate from  using v=c, where c is the speed of light.

nm)405.0 435.5 480.0 520.0 577.7650.0

V (volt)1.4751.2681.027 0.886 0.6670.381

(x1014 sec-1) 7.4026.8846.2455.7655.1894.612

1.60From Equation 1.17:

.

For the hydrogen atom the mass of the nucleus is just equal to the mass of a proton.

For the helium-4 isotope the mass of the nucleus is equal to 6.64610-27 kg.

That the percent change for a He+ is smaller illustrates that the bigger the difference in the two masses (the mass of the electron and of the nucleus), the closer the reduced mass is to the lighter mass (the mass of the electron).

1.61(a)For an s orbital, is zero. Hence, the orbital angular momentum for an electron in an s orbital is zero.

(b)Bohr’s model only allowed for orbits with angular momentum equal to an integer multiple of, whereis defined as Plank’s constant divided by two times pi 

1.62The Fe14+ in the corona is presumed to be formed by the ionization of Fe13+, which takes 3.5104 kJ mol-1 of energy. The energy for the ionization comes from collisions. Hence, for there to be Fe14+ in the corona, we can assume that the average kinetic energy of the corona gases has to be atleast equal to the ionization energy of Fe13+.

The actual temperature can be, and most probably is, higher than this.

1.63Since the energy corresponding to a photon of wavelength 1 equals the energy of photon of wavelength 2 plus the energy of photon of wavelength 3, then the equation must relate the wavelength to energy.

energy of photon 1  (energy of photon 2  energy of photon 3)

Since, then: