Chemistry 101 - 2001 Answers Assignment #2 and Quiz 2

 

Based on the following questions taken from the 3rd edition of McMurry and Fay (5.41 means question 41 from the end of chapter 5)

5.41, 5.43, 5.44, 5.45, 5.47, 5.49, 5.57, 5.63

Or based on the following questions taken from the 2nd edition of McMurry and Fay (5.41 means question 41 from the end of chapter 5)

5.45, 5.47, 5.48, 5.49, 5.51, 5.53, 5.63, 5.69

 

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3rd edtion (2nd edition)

5.41 (5.45). What is the de Broglie wavelength (in meters) of a mosquito weighing 1.55 mg and flying at 1.38 m s-1. Answer:

= h / (mv) = (6.626 x 10-34 J s) / ( 1.55 x 10-3 g x 1.38 m s-1) =

=(6.626 x 10-34 kg m2 s-2. s) / ( 1.55 x 10-6 kg x 1.38 m s-1) = 3.10 x 10-28 m.

Note: be careful in the conversion of units.

1 mg = 1 x 10-3 g = 1x 10-6 kg and 1 J = 1 kg m2 s-2.

5.43 (5.47). What velocity would an electron (mass = 9.11 x 10-31 kg) need for its de Broglie wavelength to be that of red light (750 nm)?

Answer:

v= h / (m) = (6.626 x 10-34 J s) / ( 9.11 x 10-31 kg x 750 x 10-9 m) = 9.70 x 102 m.

5.44 (5.48). According to the equation for the Balmer line spectrum of hydrogen, a value of n = 3 gives a red spectral line at 656.3 nm, a value of n = 4 gives a green line at 486.1 nm and a value of n = 5 gives a blue line at 434.0 nm. Calculate the energy (in kilojoules per mole) of the radiation corresponding to each of these spectral lines.

Answer: To get the answer in kJ mol-1 you must multiply h by NA.

a N hN hc x mol x J s x

AA)

. . .

.

E m s

656.3nm 10 m nm

J kJ mol

-1

-9 -1

-1

6 022 10 6 626 10 2 998 10

182300 182 3

23 1 34 8

b N hN hc x mol x J s x

AA)

. . .

.

E m s

486.1nm 10 m nm

J kJ mol

-1

-9 -1

-1

6 022 10 6 626 10 2 998 10

246100 246 1

23 1 34 8

c N hN hc x mol x J s x

AA)

. . .

.

E m s

434.0 nm 10 m nm

J kJ mol

-1

-9 -1

-1

6 022 10 6 626 10 2 998 10

275600 275 6

23 1 34 8

5.45 (5.49). According to the values cited in the previous question, the wavelength differences between lines in the Balmer series become smaller as n becomes larger. In other words, the wavelengths converge toward a minimum value as n becomes very large. At what wavelength (in nanometers) do the lines converge?

Answer: Balmer series m = nlower = 2; n = nhigher = infinity at the convergence limit.

R = 1.097 x 10-2 nm-1

1 1 11 097 10

1

2

12 74 10 3652 2

2 12 2

3 1

R

m nx nm x nm nm. .

infinty

5.47 (5.51). Lines in the Brackett series of the hydrogen spectrum are caused by the emission of energy accompanying the fall of an electron from outer shells to the fourth shell. The lines can be calculated using the Balmer-Rydberg equation:

1 1 12 2

Rm n

where m = 4, R = 1.097 x 10-2 nm-1 and n is an integer greater than 4. Calculate the wavelengths (in nanometers) and the energies (in kilojoules per mole) of the first two lines of the Brackett series. In what region of the electromagnetic spectrum do they fall?Answer:

Brackett series m = nlower = 4; for the first two lines n = nupper = 5 and 6.

1 1 11097 10

1

4

1

50 000247

4 05 10

6 626 101998 10

4 05 104 90 10

4 90 10 6 022 10 29 500 29

2 22 1

2 21

6

348 1

620

20 23 1 1

Rm n

x nm nm

x m

E hc

x J sx m s

x mx J

E E N x J x mol JmolA

. .

.

..

..

' . . , .

per molecule

per mole

5 1kJmol

1 1 11097 10

1

4

1

60 000381

2 63 10

6 626 101998 10

2 63 107 55 10

4 90 10 6 022 10 29 500 45

2 22 1

2 21

6

348 1

620

20 23 1 1

Rm n

x nm nm

x m

E hc

x J sx m s

x mx J

E E N x J x mol JmolA

. .

.

..

..

' . . , .

per molecule

per mole

5 1kJmol

5.49 (5.53). Excited rubidium atoms emit red light with = 795 nm. What is the energy difference (in kilojoules per mole) between orbitals that give rise to this emission?Answer:

per mole

795 7 95 10

6 626 102 998 10

7 95 106 022 10

150 000 150

7

348 1

723 1

1 1

nm x m

E E N hc

N x J sx m s

x mx mol

Jmol kJmol

A A

.

' ..

..

,

5.57 (5.63). Give orbital designations of electrons with the following quantum numbers:

a) n = 3, l = 0, ml = 0 b) n = 2, l = 1, ml = -1 c) n = 4, l = 3, ml = -2 d) n = 4, l = 2, ml = 0

Answer:

a) n = 3, l = 0, ml = 0 is a 3s orbital

b) n = 2, l = 1, ml = -1 is a 2p orbital (actually a 2p-1 orbital)

c) n = 4, l = 3, ml = -2 is a 4f orbital (actually a 4f-2 orbital)

d) n = 4, l = 2, ml = 0 is a 4d orbital (actually a 4d0 orbital)

5.63 (5.69). The mass of a helium atom is 4.0026 amu, and its average velocity at 25oC is 1.36x103 m s-1. What is the uncertainty (in meters) in the position of a helium at if the uncertainty in its velocity is 1%.

Answer:

m amu . x k

kg ms x kg ms

x kg ms x kg ms

x mh

xh

m

x kg m s

x kg msx m nm

He

4 0026 1661 10

9 04 10

9 04 10 9 04 10

4

4

6 626 10

4 31415926 9 04 10583 10 0583

27

1 24 1

24 1 26 1

34 2 1

26 110

.

.

. .

.

. .. .

x g = 6.648x10 kg

mv = 6.648x10 x 1.36 x 10

mv = 0.01 x

v

v

-27

-27 3

Thus, the uncertainty in measuring the position of the He atom is comparable to the size of the He atom (whose diameter is approximately 0.2 nm)

Quiz questions not taken directly from assignment  

Section F01 ( quizzes A*, C*, E*)Section F03 ( quizzes B”, D”) 8. Which of the following will have the LONGEST de Broglie wavelength?

a) an electron with a velocity of 1.0x106 m s-1 b) a proton with a velocity of 1.0x106 m s-1 c) an electron with a velocity of 2.0x106 m s-1 d) a proton with a velocity of 2.0x106 m s-1 e) a neutron with a velocity of 1.0x106 m s-1

Answer:

de Broglie s equation' : is =h

mv

For to be large we want m and v both to be small. Since the mass of the electron is much smaller (by a factor of 2000) than the proton or neutron, the correct answer must be a) in the list above since in c) the electron is moving faster that in a). 9. If the value of the Rydberg constant is 1.097x10-2 nm-1, the wavelength of the SECOND line in the Lyman seris (nlower = 1) is: 

Answer: The second line in the Lyman series is n = 3 going to n = 1. Substituting this in the equation we get: 1 1 1

1 097 101

1

1

39 75 10

1

9 75 10102 6

2 22 1

2 23 1

3 1

Rn n

x nm x nm

x nmnm

lower upper

. .

..

 10. If {n,l,ml} represents the orbital specified by n=n, l=l and ml=ml, then which of the following is an INCORRECT designation for the specified orbital? Answer: The first number is the n value in both representations so they have to match. The second number is the l value where l=0 is an s orbital, l=1 is a p orbital, l=2 is a d orbital and l=3 is an f orbital. The third number ml has to integral and be between -l and +l .  In this question all the matches are correct.

  

 Section F01( quizzes B*, D*)Section F02 ( quizzes A, B, C, D, E)Section F03( quizzes A”, C”, E”) 8. Which of the following will have the SHORTEST de Broglie wavelength?

a) an electron with a velocity of 1.0x106 m s-1 b) a proton with a velocity of 1.0x106 m s-1 c) an electron with a velocity of 2.0x106 m s-1 d) a proton with a velocity of 2.0x106 m s-1 e) a neutron with a velocity of 1.0x106 m s-1

Answer:

de Broglie s equation' : is =h

mv

For to be small we want m and v both to be large. Since the mass of the electron is much smaller (by a factor of 2000) than the proton or neutron, we can eliminate the electron. The masses of the proton and

neutron are almost the same, so correct answer must be d) in the list above since in b) and e) the velocity is only half that in d).

 9. If the value of the Rydberg constant is 1.097x10-2 nm-1, the wavelength of the THIRD line in the Lyman seris (nlower = 1) is: Answer: The third line in the Lyman series is n = 4 going to n = 1. Substituting this in the equation we get: 1 1 1

1 097 101

1

1

41 028 10

1

1 028 1097 2

2 22 1

2 22 1

2 1

Rn n

x nm x nm

x nmnm

lower upper

. .

..

 10. If {n,l,ml} represents the orbital specified by n=n, l=l and ml=ml, then which of the following is an INCORRECT designation for the specified orbital?  Answer: The first number is the n value in both representations so they have to match. The second number is the l value where l=0 is an s orbital, l=1 is a p orbital, l=2 is a d orbital and l=3 is an f orbital. The third number ml has to integral and be between -l and +l . Therefore the designation {5,4,1} is not a 5f orbital but a 5g orbital since l is not 3.

 Section F04 ( quizzes A’, B’, C’, D’, E) 8. Which of the following will have the SHORTEST de Broglie wavelength?

a) an electron with a velocity of 1.0x106 m s-1 b) a proton with a velocity of 1.0x106 m s-1 c) an electron with a velocity of 2.0x106 m s-1 d) a proton with a velocity of 2.0x106 m s-1 e) a neutron with a velocity of 1.0x106 m s-1

Answer:

de Broglie s equation' : is =h

mv

For to be small we want m and v both to be large. Since the mass of the electron is much smaller (by a factor of 2000) than the proton or neutron, we can eliminate the electron. The masses of the proton and neutron are almost the same, so correct answer must be d) in the list above since in b) and e) the velocity is only half that in d).

 9. If the value of the Rydberg constant is 1.097x10-2 nm-1, the wavelength of the SECOND line in the Lyman seris (nlower = 1) is: Answer: The second line in the Lyman series is n = 3 going to n = 1. Substituting this in the equation we get: 1 1 1

1 097 101

1

1

39 75 10

1

9 75 10102 6

2 22 1

2 23 1

3 1

Rn n

x nm x nm

x nmnm

lower upper

. .

..

 10. If {n,l,ml} represents the orbital specified by n=n, l=l and ml=ml, then which of the following is an INCORRECT designation for the specified orbital? Answer: The first number is the n value in both representations so they have to match. The second number is the l value where l=0 is an s orbital, l=1 is a p orbital, l=2 is a d orbital and l=3 is an f orbital. The third number ml has to integral and be between -l and +l .  In this question all the matches are correct.