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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.)
1.4 The 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 = hn
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 = hn = (6.63 ´ 10-34 J s)(6.0 ´ 104 s-1) = 4.0 ´ 10-29 J
(c) Converting to J mol-1:
1.8 The energy given in this problem is for 1 mole of photons. To apply E = hn, 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 = hn. Substitute the frequency (n) into this equation to solve for the energy of one quantum associated with this frequency.
1.11 First, 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:
1.12 The 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:
1.13 This 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:
DE = (3.372 ´ 10-19 J) - (3.369 ´ 10-19 J) = 3 ´ 10-22 J
For one mole of photons the energy difference is:
1.14 The radiation emitted by the star is measured as a function of wavelength (l). The wavelength corresponding to the maximum intensity is then plugged in to Wien’s law (Tlmax = 1.44 ´ 10-2 K m) to determine the surface temperature.
1.15 Wien’s law: Tlmax = 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.
1.16 Because 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.
1.17 In 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
1.20 The emitted light could be analyzed by passing it through a prism.
1.21 Light 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.
1.22 Excited 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 DE, the wavelength is found as in part (a).
1.24 We use more accurate values of h and c for this problem.
1.25 In this problem ni = 3 and nf = 5.
The sign of DE 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?
1.26 Strategy: 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:
DE = -4.09 ´ 10-19 J
The negative sign for DE indicates that this is energy associated with an emission process. To calculate the frequency, we will omit the minus sign for DE because the frequency of the photon must be positive. We know that
DE = hn
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).
1.27 The Balmer series corresponds to transitions to the n = 2 level.
For He+:
For the transition, n = 6 ® 2
For the transition, n = 6 ® 2, DE = -1.94 ´ 10-18 J l = 103 nm
For the transition, n = 5 ® 2, DE = -1.83 ´ 10-18 J l = 109 nm
For the transition, n = 4 ® 2, DE = -1.64 ´ 10-18 J l = 121 nm
For the transition, n = 3 ® 2, DE = -1.21 ´ 10-18 J l = 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, DE = -4.84 ´ 10-19 J l = 411 nm
For the transition, n = 5 ® 2, DE = -4.58 ´ 10-19 J l = 434 nm
For the transition, n = 4 ® 2, DE = -4.09 ´ 10-19 J l = 487 nm
For the transition, n = 3 ® 2, DE = -3.03 ´ 10-19 J l = 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.
1.28 Equation 1.14:
for n = 2:
for n = 3:
1.29 ni = 236, nf = 235
This wavelength is in the microwave region. (See Figure 1.6 of the text.)
1.30
1.31 First convert the parameters to the appropriate units and then use the following equation:
(a)
(b)
1.32
1.33 Strategy: 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.
1.34 Converting the velocity to units of ms-1:
1.35 First, we convert mph to m s-1.
1.36 From 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 Dp = m Du, the uncertainty in the velocity is given by
The minimum uncertainty in the velocity is 1 ´ 106 m s–1.
1.37 Below 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.
1.38 The 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 1 Å=10-10m.
Energy is equal to Planck’s constant times frequency. We can rearrange this equation to solve for frequency.
1.39 From 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