Silberberg7e solution manual ch 07

nfinal = 2 7.4A Plan: With the equation for the de Broglie wavelength,  = h/mu and the given de Broglie wavelength, calculate the electron speed.. 7.2 Plan: Recall that the shorter the

CHAPTER 7 QUANTUM THEORY AND

wavelength must be in units of m in this equation

7.2A Plan: Use the formula E = hto solve for the frequencyThen use the equation c = to solve for the wavelength

Solution:

 = E

h=

8.2x10–19 J 6.626x10–34 J•s = 1.2375x1015 = 1.2x10 15 s –1

(using the unrounded number in the next calculation to avoid rounding errors)

7.3A Plan: Use the equation relating ΔE = – 2.18x10– 18 J 2 2

n n  to find the energy change; a photon in

the IR (infrared) region is emitted when has n has a final value of 3 Then use E = hc/to find the wavelength of

nfinal2 = 0.25025

nfinal2 = 3.9960 = 4 (The final energy level is an integer, so its square is also an integer.)

nfinal = 2

7.4A Plan: With the equation for the de Broglie wavelength,  = h/mu and the given de Broglie wavelength, calculate

the electron speed The wavelength must be expressed in meters Use the same formulas to calculate the speed of the golf ball The mass of both the electron and the golf ball must be expressed in kg in their respective

calculations

6.626 x10 J • s kg • m /s

J

10 m9.11x10 kg 100.nm

kg•m2/s2

J = 1.4436x10–25 = 1.44x10 –25 m/s

7.4B Plan: Use the equation for the de Broglie wavelength,  = h/mu with the given mass and speed to calculate the de

Broglie wavelength of the racquetball The mass of the racquetball must be expressed in kg, and the speed must be expressed in m/s in the equation for the de Broglie wavelength

Solution:

Mass (kg) of the racquetball = (39.7 g) 1 kg

1000 g = 0.0397 kg Speed (m/s) of the racquetball = 55 mi

hr

1 hr

3600 s

1.609 km mi

7.5A Plan: Use the equation •

346.626x10 J • s

πm u =

6.626x10–34 J•s 4π 1.67x10–27 kg (8x105m/s) = 3.9467x10– 14 m

For l = 1, m l = –1, 0, 1

For l = 2, m l = –2, –1, 0, 1, 2

For l = 3, m l = –3, –2, –1, 0, 1, 2, 3

7.6B Plan: Following the rules for l (integer from 0 to n – 1) and m l (integer from –l to +l), determine which value of

the principal quantum number has five allowed levels of l

7.7A Plan: Identify n and l from the sublevel designation n is the integer in front of the sublevel letter The sublevels

are given a letter designation, in which s represents l = 0, p represents l = 1, d represents l = 2, f represents l = 3 Knowing the value for l, find the m l values (integer from –l to +l)

Solution:

Sublevel name n value l value m l values

7.7B Plan: Identify n and l from the sublevel designation n is the integer in front of the sublevel letter The sublevels

are given a letter designation, in which s represents l = 0, p represents l = 1, d represents l = 2, f represents l = 3 Knowing the value for l, find the m l values (integer from –l to +l)

7.8A Plan: Use the rules for designating quantum numbers to fill in the blanks

For a given n, l can be any integer from 0 to n–1

For a given l, m l can be any integer from – l to + l

The sublevels are given a letter designation, in which s represents l = 0, p represents l = 1, d represents l = 2,

7.8B Plan: Use the rules for designating quantum numbers to determine what is wrong with the quantum number

designations provided in the problem

For a given n, l can be any integer from 0 to n–1

For a given l, m l can be any integer from – l to + l

The sublevels are given a letter designation, in which s represents l = 0, p represents l = 1, d represents l = 2,

f represents l = 3

Solution:

The provided table is:

a) 5 3 4 5f b) 2 2 1 2d c) 6 1 –1 6s

a) For l = 3, the allowed values for m l are –3, –2, –1, 0, 1, 2, 3, not 4

b) For n = 2, l = 0 or 1 only, not 2; the sublevel is 2p, since m l = 1

c) The value l = 1 indicates the p sublevel, not the s; the sublevel name is 6p

TOOLS OF THE LABORATORY BOXED READING PROBLEMS

B7.1 Plan: Plot absorbance on the y-axis and concentration on the x-axis Since this is a linear plot, the graph is of the

type y = mx + b, with m = slope and b = intercept Any two points may be used to find the slope, and the slope is

used to find the intercept Once the equation for the line is known, the absorbance of the solution in part b) is used to find the concentration of the diluted solution, after which the dilution equation is used to find the

molarity of the original solution



4/M

Using the slope just calculated and any of the data points, the value of the intercept may be found

b = y – mx = 0.396 – (13,250/M)(3.0x10–5 M) = –0.0015 = 0.00 (absorbance has no units)

b) Use the equation just determined: y = (1.3x104/M) x + 0.00

Concentration (M)

frequency of each wavelength can be determined from the relationship c = or =



c

The wavelength in nm must be converted to meters

7.1 All types of electromagnetic radiation travel as waves at the same speed They differ in both their frequency,

wavelength, and energy

7.2 Plan: Recall that the shorter the wavelength, the higher the frequency and the greater the energy Figure 7.3

describes the electromagnetic spectrum by wavelength and frequency

Solution:

a) Wavelength increases from left (10–2 nm) to right (1012 nm) in Figure 7.3 The trend in increasing wavelength

is: x-ray < ultraviolet < visible < infrared < microwave < radio wave

b) Frequency is inversely proportional to wavelength according to the equation c = λν, so frequency has the

opposite trend: radio wave < microwave < infrared < visible < ultraviolet < x-ray

c) Energy is directly proportional to frequency according to the equation E = hν Therefore, the trend in increasing

energy matches the trend in increasing frequency: radio wave < microwave < infrared < visible < ultraviolet < x-ray

7.3 a) Refraction is the bending of light waves at the boundary of two media, as when light travels from air into water

b) Diffraction is the bending of light waves around an object, as when a wave passes through a slit about as wide as

c) Dispersion is the separation of light into its component colors (wavelengths), as when light passes through a prism

d) Interference is the bending of light through a series of parallel slits to produce a diffraction pattern of

brighter and darker spots

Note: Refraction leads to a dispersion effect and diffraction leads to an interference effect

7.4 Evidence for the wave model is seen in the phenomena of diffraction and refraction Evidence for the particle model

includes the photoelectric effect and blackbody radiation

7.5 a) Frequency: C < B < A

b) Energy: C < B < A

c) Amplitude: B < C < A

d) Since wave A has a higher energy and frequency than B, wave A is more likely to cause a current

e) Wave C is more likely to be infrared radiation since wave C has a longer wavelength than B

7.6 Radiation (light energy) occurs as quanta of electromagnetic radiation, where each packet of energy is called a

photon The energy associated with this photon is fixed by its frequency, E = hν Since energy depends on frequency,

a threshold (minimum) frequency is to be expected A current will flow as soon as a photon of sufficient energy reaches the metal plate, so there is no time lag

7.7 Plan: Wavelength is related to frequency through the equation c = ν Recall that a Hz is a reciprocal second, or

7.11 Plan: Energy is inversely proportional to wavelength ( = E hc

 ) As wavelength decreases, energy increases

Solution:

In terms of increasing energy the order is red < yellow < blue

7.12 Since energy is directly proportional to frequency (E = h): UV ( = 8.0x1015 s–1) > IR ( = 6.5x1013 s–1) > microwave ( = 9.8x1011 s–1) or UV > IR > microwave

7.13 Plan: Wavelength is related to frequency through the equation c = ν Recall that a Hz is a reciprocal second,

  = 1.3482797x10

7 = 1.3483x10 7 nm

2.9979x10 m/s 1 m

10 ms

The wavelength can also be found using the frequency calculated in the equation c = 

7.16 Plan: The least energetic photon in part a) has the longest wavelength (242 nm) The most energetic photon in

part b) has the shortest wavelength (2200 Å) Use the relationship c = to find the frequency of the photons and

7.17 “n” in the Rydberg equation is equal to a Bohr orbit of quantum number “n” where n = 1, 2, 3, 

7.18 Bohr’s key assumption was that the electron in an atom does not radiate energy while in a stationary state, and the

electron can move to a different orbit by absorbing or emitting a photon whose energy is equal to the difference in energy between two states These differences in energy correspond to the wavelengths in the known spectra for the hydrogen atoms A Solar System model does not allow for the movement of electrons between levels 7.19 An absorption spectrum is produced when atoms absorb certain wavelengths of incoming light as electrons move

from lower to higher energy levels and results in dark lines against a bright background An emission spectrum is

produced when atoms that have been excited to higher energy emit photons as their electrons return to lower energy levels and results in colored lines against a dark background Bohr worked with emission spectra

7.20 Plan: The quantum number n is related to the energy level of the electron An electron absorbs energy to

change from lower energy (lower n) to higher energy (higher n), giving an absorption spectrum An electron emits energy as it drops from a higher energy level (higher n) to a lower one (lower n), giving an emission spectrum

Solution:

a) The electron is moving from a lower value of n (2) to a higher value of n (4): absorption

b) The electron is moving from a higher value of n (3) to a lower value of n (1): emission

c) The electron is moving from a higher value of n (5) to a lower value of n (2):emission

d) The electron is moving from a lower value of n (3) to a higher value of n (4): absorption

7.21 The Bohr model works only for a one-electron system The additional attractions and repulsions in many-electron

systems make it impossible to predict accurately the spectral lines

7.22 The Bohr model has successfully predicted the line spectra for the H atom and Be3+ ion since both are

one-electron species The energies could be predicted from En =  2 18 

22.18x10 J

Z n





where Z is the atomic number

for the atom or ion The line spectra for H would not match the line spectra for Be3+ since the H nucleus contains one proton while the Be3+ nucleus contains 4 protons (the Z values in the equation do not match); the force of

attraction of the nucleus for the electron would be greater in the beryllium ion than in the hydrogen atom This means that the pattern of lines would be similar, but at different wavelengths

7.23 Plan: Calculate wavelength by substituting the given values into Equation 7.3, where n1 = 2 and n2 = 5 because

n2 > n1 Although more significant figures could be used, five significant figures are adequate for this calculation Solution:

7.24 Calculate wavelength by substituting the given values into the Rydberg equation, where n1 = 1 and n2 = 3 because

n2 > n1 Although more significant figures could be used, five significant figures are adequate for this calculation

7.25 Plan: The Rydberg equation is needed For the infrared series of the H atom, n1 equals 3 The least energetic

spectral line in this series would represent an electron moving from the next highest energy level, n2 = 4 Although more significant figures could be used, five significant figures are adequate for this calculation

7.26 Plan: The Rydberg equation is needed For the visible series of the H atom, n1 equals 2 The least energetic

spectral line in this series would represent an electron moving from the next highest energy level, n = 3 Although

more significant figures could be used, five significant figures are adequate for this calculation

7.27 Plan: To find the transition energy, use the equation for the energy of an electron transition and multiply by

Avogadro’s number to convert to energy per mole

The energy has a negative value since this electron transition to a lower n value is an emission of energy

7.28 To find the transition energy, use the equation for the energy of an electron transition and multiply by Avogadro’s

Frequency is proportional to energy so the smallest frequency will be d) n = 4 to n = 3; levels 3 and 4 have a

smaller E than the levels in the other transitions The largest frequency is b) n = 2 to n = 1 since levels 1 and

2 have a larger E than the levels in the other transitions Transition a) n = 2 to n = 4 will be smaller than

transition c) n = 2 to n = 5 since level 5 is a higher energy than level 4 In order of increasing frequency the

transitions are d < a < c < b

7.30 b > c > a > d

7.31 Plan: Use the Rydberg equation Since the electron is in the ground state (lowest energy level), n1 = 1 Convert

the wavelength from nm to units of meters

7.35 If an electron occupies a circular orbit, only integral numbers of wavelengths (= 2nr) are allowed for acceptable

standing waves A wave with a fractional number of wavelengths is forbidden due to destructive interference with itself In a musical analogy to electron waves, the only acceptable guitar string wavelengths are those that are an integral multiple of twice the guitar string length (2 L)

7.36 De Broglie’s concept is supported by the diffraction properties of electrons demonstrated in an electron

microscope

7.37 Macroscopic objects have significant mass A large m in the denominator of  = h/mu will result in a very small

wavelength Macroscopic objects do exhibit a wavelike motion, but the wavelength is too small for humans to see

it

7.38 The Heisenberg uncertainty principle states that there is fundamental limit to the accuracy of measurements This

limit is not dependent on the precision of the measuring instruments, but is inherent in nature

7.39 Plan: Use the de Broglie equation Mass in lb must be converted to kg and velocity in mi/h must

be converted to m/s because a joule is equivalent to kg•m2/s2

7.41 Plan: Use the de Broglie equation Mass in g must be converted to kg and wavelength in Ǻ must be converted to

m because a joule is equivalent to kg•m2/s2

7.43 Plan: The de Broglie wavelength equation will give the mass equivalent of a photon with known wavelength and

velocity The term “mass equivalent” is used instead of “mass of photon” because photons are quanta of

electromagnetic energy that have no mass A light photon’s velocity is the speed of light, 3.00x108 m/s

Wavelength in nm must be converted to m

7.45 The quantity 2 expresses the probability of finding an electron within a specified tiny region of space

7.46 Since 2 is the probability of finding an electron within a small region or volume, electron density would

represent a probability per unit volume and would more accurately be called electron probability density

7.47 A peak in the radial probability distribution at a certain distance means that the total probability of finding the

electron is greatest within a thin spherical volume having a radius very close to that distance Since principal

quantum number (n) correlates with distance from the nucleus, the peak for n = 2 would occur at a greater

distance from the nucleus than 0.529 Å Thus, the probability of finding an electron at 0.529 Å is much greater for

the 1s orbital than for the 2s

7.48 a) Principal quantum number, n, relates to the size of the orbital More specifically, it relates to the distance from

the nucleus at which the probability of finding an electron is greatest This distance is determined by the energy of

b) Angular momentum quantum number, l, relates to the shape of the orbital It is also called the azimuthal

c) Magnetic quantum number, m l, relates to the orientation of the orbital in space in three-dimensional space

7.49 Plan: The following letter designations correlate with the following l quantum numbers:

l = 0 = s orbital; l = 1 = p orbital; l = 2 = d orbital; l = 3 = f orbital Remember that allowed m l values are – l to + l The number of orbitals of a particular type is given by the number of possible m l values

Solution:

a) There is only a single s orbital in any shell l = 1 and m l = 0: one value of m l = one s orbital

b) There are five d orbitals in any shell l = 2 and m l = –2, –1, 0, +1, +2 Five values of m l = five d orbitals

c) There are three p orbitals in any shell l = 1 and m l = –1, 0, +1 Three values of m l = three p orbitals

d) If n = 3, l = 0(s), 1(p), and 2(d) There is a 3s (1 orbital), a 3p set (3 orbitals), and a 3d set (5 orbitals) for a

total of nine orbitals (1 + 3 + 5 = 9)

7.50 a) All f orbitals consist of sets of seven (l = 3 and m l = –3, –2, –1, 0, +1, +2, +3)

b) All p orbitals consist of sets of three (l = 1 and m l = –1, 0, +1)

c) All d orbitals consist of sets of five (l = 2 and m l = –2, –1, 0, +1, +2)

d) If n = 2, then there is a 2s (1 orbital) and a 2p set (3 orbitals) for a total of four orbitals (1 + 3 = 4)

7.51 Plan: Magnetic quantum numbers (m l ) can have integer values from –l to + l The l quantum number can

have integer values from 0 to n – 1

Solution:

a) l = 2 so m l = –2, –1, 0, +1, +2

b) n = 1 so l = 1 – 1 = 0 and m l = 0

c) l = 3 so m l = –3, –2, –1, 0, +1, +2, +3

7.52 Magnetic quantum numbers can have integer values from –l to +l The l quantum number can

have integer values from 0 to n – 1

a) l = 3 so m l = –3, –2, –1, 0, +1, +2, +3

b) n = 2 so l = 0 or 1; for l = 0, m l = 0; for l = l, m l = –1,0,+1

7.53 Plan: The s orbital is spherical; p orbitals have two lobes; the subscript x indicates that this orbital lies along

The variations in coloring of the p orbital are a consequence of the quantum mechanical derivation of atomic

orbitals that are beyond the scope of this course

7.54 a) p z : 2 lobes along the z-axis b) d xy: 4 lobes

xy

z

xy

The variations in coloring of the p and d orbitals are a consequence of the quantum mechanical derivation of

atomic orbitals that are beyond the scope of this course

7.55 Plan: The following letter designations for the various sublevels (orbitals) correlate with the following l quantum

numbers: l = 0 = s orbital; l = 1 = p orbital; l = 2 = d orbital; l = 3 = f orbital Remember that allowed m l values are

– l to + l The number of orbitals of a particular type is given by the number of possible m l values.

7.57 Plan: The integer in front of the letter represents the n value The letter designates the l value:

l = 0 = s orbital; l = 1 = p orbital; l = 2 = d orbital; l = 3 = f orbital Remember that allowed m l values are – l to + l

Solution:

a) For the 5s subshell, n = 5 and l = 0 Since m l = 0, there is one orbital