Practice Problems 13
Chapter 7
CHE 151 Graham/07
1.)
A laser emits light of frequency 4.74 x 1014 sec-1. What is the wavelength of the light in nm?
= c = 2.998 x 108 m x 1 s x 1 nm = 6.32 x 102 nm s 4.74 x 1014 10-9m
2.)
A certain electromagnetic wave has a wavelength of 625 nm. a.) What is the frequency of the wave?
625 nm x 10-9 m = 6.25 x 10-7 m 1 nm
= c = 2.998 x 10 m/s 6.25 x 10-7 m 8
= 4.80 x 1014 s-1
b.) What region of the electromagnetic spectrum is it found? Visible Region (~400 – 750 nm)
c.) What is the energy of the wave? E = h= (6.626 x 10-34 J.s)(4.80 x 1014 s-1) = 3.18 x 10-19 J
3.)
How many minutes would it take a radio wave to travel from the planet Venus to Earth? (Average distance from Venus to Earth = 28 million miles). (Note: All electromagnetic travels at the speed of light in a vacuum) 2.8 x 107 mi 4.5 x 1010 m
4.)
x 1 km x 0.6214 mi
103 m = 4.5 x 1010 m 1 km
x 1s x 8 2.998 x 10 m
1 min = 60 s
2.5 min
The blue color of the sky results from the scattering of sunlight by air molecules. The blue light has a frequency of about 7.5 x 1014 Hz. a.) Calculate the wavelength, in nm, associated with this radiation. = c = 2.998 x 10 m x 1 nm = 4.0 x 10 nm 7.5 x 1014 s-1 10-9 m 8
1 Hz = 1 s-1
2
b.) Calculate the energy, in joules, of a single photon associated with this frequency. E = h= (6.626 x 10-34 J.s)(7.5 x 1014 s-1) = 5.0 x 10-19 J
What is E in joules for an atom that releases a photon with a wavelength of 3.2 x 10-7 meters? Eatom = Ephoton = hhc
5.)
E = (6.626 x 10-34 J.s)( 2.998 x 108 m/s) = 6.2 x 10-19 J 3.2 x 10-7 m
6.)
Calculate the frequency (Hz) and wavelength (nm) of the emitted photon when an electron drops from the n=4 to n=2 state. E = RH 1 - 1 = (2.179 x 10-18J) 1 - 1 = 2.179 x 10-18 - 2.179 x 10-18 ni2 nf2 42 22 16 4
E = h = E
= 1.362 x 10-19 - 5.448 x 10-19 = -4.086 x 10-19J = E = 4.086 x 10-19J = 6.167 x 1014Hz = 6.626 x 10-34 J.s
h
= c =
7.)
2.998 x 108 m/s x 1 nm 6.167 x 1014 s-1 10-9m
=
486.1 nm
An electron in the hydrogen atom makes a transition from an energy state of principal quantum numbers ni to the n = 2 state. If the photon emitted has a wavelength of 434 nm, what is the value of ni?
Eatom = Ephoton = hhc = (6.626 x 10-34 J.s)( 2.998 x 108 m/s) = -4.58 x 10-19 J 434 x 10-9m (negative number because it is an emission process)
E = RH 1 - 1 = -4.58 x 10-19J = (2.179 x 10-18J) ni2 nf2 -4.58 x 10-19J = 1 - 0.250 (keep 3 sf) 2.179 x 10-18J ni2 ni =
8.)
1 - 1 ni2 22
-0.210 + 0.250 = 1 = 0.040
1 √ 0.040
ni2
ni = 5
Protons can be accelerated to speeds near that of light in particle accelerators. Estimate the deBroglie wavelength (in nm) of such a proton moving at 2.90 x 108 m/s. (mass of a proton = 1.673 x 10-27 kg).
1 J = 1 kg . m2 s2 =h mu
h = 6.626 x 10-34 J.s x 1 kg . m2 = 6.626 x 10-34 kg . m2 s2 s =
6.626 x 10-34 kg . m2/s = 1.37 x 10-15 m -27 8 (1.673 x 10 kg)(2.90 x 10 m/s)
1.37 x 10-15 m x 1nm = 1.37 x 10-6 nm 10-9m
9.)
Calculate the deBroglie wavelength (in nm) of a 3000. lb automobile traveling at 55 mi/hr. 55 mi x 1 km x 103 m x 1 hr x 1 min = 25 m 1 hr 0.6214 mi 1 km 60 min 60 sec s
3000. lb x 1 kg = 1361 kg 2.2046 lb
= h = 6.626 x 10-34 kg . m2/s = 1.9 x 10-38 m x 1 nm = 1.9 x 10-29 nm mu (1361 kg)(25 m/s) 10-9 m
10.)
What are the possible values of l for an electron with n=3? l = (0….n-1)
11.)
l = 0,1,2
For the following subshells give the values of the quantum numbers (n, l and ml) and the number of orbitals in each subshell. (a) 4p n=4 l=1 ml = -1,0,+1
(b) 3d n=3 l =2 ml = -2,-1,0,+1,+2
(c) 3s n=3 l=0 ml = 0
(3 p orbitals)
(5 d orbitals)
(1 s orbital)
(d) 5f n=5 l=3 ml = -3,-2,-1, 0,+1,+2,+3 (7 f orbitals)
12.)
13.)
For each of the following, give the subshell designation, the allowable ml values, and the number of orbitals. (a) n = 2, l = 0
2s, ml = 0, (1 orbital)
(b) n = 3, l = 2
3d, ml = -2,-1,0,+1,+2 (5 orbitals)
(c) n = 5, l = 1
5p, ml = -1,0,+1
(3 orbitals)
Are the following quantum number combinations allowed? If not, show two ways to correct them. (a) n = 1; l = 0; ml = 0 yes. 1s (b) n = 2; l = 2; ml = +1
No n = 3; l = 2; ml = +1 or n = 2; l = 1; ml = +1
(c) n = 7; l = 1; ml = +2 n = 7; l = 1; ml = +1
(d) n = 3; l = 1; ml = -2 n = 3; l = 1; ml = -1
No or n = 7; l = 2; ml = +2 No or
n = 3; l = 2; ml = -2
14.)
The energy required to remove an electron from metal X is E = 3.31 x 10-20J. Calculate the maximum wavelength of light that can photo eject an electron from metal X.
E = hc
= hc = (6.626 x 10-34 J.s)( 2.998 x 108 m/s) E 3.31 x 10-20 J = 6.00 x 10-6 m x 1nm = 10-9m
15.)
If an electron has a velocity of 5.0 x 105 m/s, what is its wavelength in m?
1 J = 1 kg . m2 s2 =h mu
m = mass of electron = 9.109 x 10-28g x 1 kg = 9.109 x 10-31 kg 103 -34 . . 2 h = 6.626 x 10 J s x 1 kg m = 6.626 x 10-34 kg . m2 s2 s =
16.)
6.00 x 103 nm
6.626 x 10-34 kg . m2/s (9.109 x 10-31 kg)(5.0 x 105 m/s)
= 1.5 x 10-9 m
The laser used to read information from a compact disk has a wavelength of 780 nm. What is the energy associated with one photon of this radiation? Ephoton = hc = (6.626 x 10-34 J.s)( 2.998 x 108 m/s) = 2.55 x 10-19 J 780 x 10-9 nm
17.)
The retina of a human eye can detect light when radiant energy incident on it is at least 4.0 x 10-17 J. For light of 600 nm wavelength, how many photons does this correspond to? 1.) Determine the energy of 1 photon:
Ephoton = hc = (6.626 x 10-34 J.s)( 2.998 x 108 m/s) = 3.31 x 10-19 J/ photon 600 x 10-9 nm 2.) Calculate # photons needed to produce given amount of energy: 4.0 x 10-17 J x 1 photon = 1.2 x 102 photons 3.31 x 10-19 J
