Download Physics 43 HW 10 Ch 39

Physics 43 HW 11 Ch 39
E: 21, 22, 24, 27, 29, 30, 33
P: 47, 50, 51, 53, 56, 61, 63
39.21. Model: To conserve energy, the absorption spectrum must have exactly the energy gained by the
atom in the quantum jumps.
Visualize: Please refer to Figure EX39.20.
Solve: (a) An electron with a kinetic energy of 2.00 eV can collide with an atom in the n = 1 state and raise its
energy to the n = 2 state. This is possible because E2 – E1 = 1.50 eV is less than 2.00 eV. On the other hand, the
atom cannot be excited to the n = 3 state.
(b) The atom will absorb 1.50 eV of energy from the incoming electron, leaving the electron with 0.50 eV of kinetic
energy.
39.22. Model: To conserve energy, the emission and the absorption spectra must have exactly the energy
lost or gained by the atom in the appropriate quantum jumps.
Solve: (a)
(b) From Equation 39.4, the energy of a light quantum is E = hf = hc/λ.
We can use this equation to find the emission and absorption
wavelengths. The emission energies from the above energy-level diagram
are: E2→1 = 4.00 eV, E3→1 = 6.00 eV, and E3→2 = 2.00 eV. The
wavelength corresponding to the 2→1 transition is
λ 2→1 =
hc ( 4.14 × 10
=
E21
−15
eV s )( 3.0 × 108 m/s )
4.00 eV
= 311 nm
Likewise, λ 3→1 = hc E3→1 = 207 nm, and λ 3→2 = 622 nm.
(c) Absorption transitions start from the n = 1 ground state. The energies
=
in
the
atom’s
absorption
spectrum
are
E1→2
4.00 eV and E1→3 = 6.00 eV. The corresponding wavelengths are
λ 1→ 2 = hc E1→ 2 = 311 nm and λ1→3 = hc E1→3 = 207 nm.
39.24. Solve: The Bohr radius is defined as
−12
−34
2
2
4πε 0h 2 4π ( 8.85 × 10 C /N m )(1.05 × 10 J s )
aB =
=
= 5.26 × 10−11 m = 0.0526 nm
2
−31
−19
me2
( 9.11×10 kg )(1.60 × 10 C )
2
This differs slightly from the accepted value of 0.0529 nm because of rounding error due to using constants
accurate to only 3 significant figures. From Equation 39.29, the ground state energy level of hydrogen is
− ( 9 × 109 N 2 m/C 2 )(1.60 × 10−19 C )
−e 2
E1 =
=
= −2.18 × 10 −18 J = − 13.61 eV
4πε 0 ( 2aB )
2 ( 5.29 × 10−11 m )
2
The slight difference from the accepted −13.60 eV is due to rounding error.
39.27. Solve: From Equation 39.30, the energy of the hydrogen atom in its first excited state is
E2 =
−13.60 eV
( 2)
2
= −3.40 eV
The ionization energy of the hydrogen atom in its first excited state (n = 2) is thus 3.40 eV.
39.29. Solve: The units of h are the units of angular momentum, L = mvr. These units are
J ⋅ s = kg ⋅
m2
m
⋅ s = kg ⋅ ⋅ m
2
s
s
39.30. Solve: Photons emitted from the n = 4 state start in energy level n = 4 and undergo a quantum jump to
a lower energy level with m < 4. The possibilities are 4 → 1, 4 → 2, and 4 → 3. According to Equation 39.36,
the transition 4 → m emits a photon of wavelength.
λ0
91.18 nm
λ=
=
1
1
1 ⎞
⎛
⎞ ⎛ 1
⎜ 2− 2⎟ ⎜ 2− ⎟
n ⎠ ⎝ m 16 ⎠
⎝m
These values are given in the table below.
Transition
Wavelength
4→1
97.3 nm
4→2
486 nm
4→3
1876 nm
39.33. Solve: For hydrogen-like ions, Equation 39.37 is
rn = n 2
aB ( rn )H
=
Z
Z
vn = Z
v1
= Z ⋅ ( vn )H
n
⎛ 13.60 eV ⎞
2
En = − Z 2 ⎜
⎟ = Z ( En ) H
n2
⎝
⎠
Where (rn)H, (vn)H, and (En)H are the values of ordinary hydrogen. He+ has Z = 2. Using Table 39.2 for the values
of hydrogen, we get
n
rn (nm)
vn (m/s)
En (eV)
1
2
3
4.38 × 106
2.19 × 106
1.46 × 106
0.026
0.106
0.238
−54.4
−13.6
−6.0
39.47. Model: The energy of a confined particle in a one-dimensional box is quantized.
Solve: The energy of the nth quantum state of a particle in a box is En = n2E1, where E1 is the lowest energy
level. The energies 12 eV, 27 eV, and 48 eV have the ratios 4:9:16. Thus, they are the n = 2, n = 3, and n = 4
states of an electron that has E1 = 3 eV = 4.8 × 10−19 J. The lowest energy level is
( 6.63 ×10−34 J s )
h2
h2
E1 =
L
⇒
=
=
= 3.5 × 10−10 m = 0.35 nm
8mL2
8mE1
8 ( 9.11 × 10−31 kg )( 4.8 × 10−19 J )
2
39.50. Model: Photons are emitted when an atom undergoes a quantum jump from a higher energy level to a
lower energy level. On the other hand, photons are absorbed in a quantum jump from a lower energy level to a
higher energy level. Because most of the atoms are in the n = 1 ground state, the only quantum jumps in the
absorption spectrum start from the n = 1 state.
hc
= ΔEatom , the 3 wavelengths in the absorption spectrum give 2.49 eV, 4.14 eV,
Solve: (a) Using Ephoton =
λ
and 6.21 eV as the energies of the n = 2, 3, and 4 energy levels.
(b) The emission spectrum of the atom will contain the following wavelengths:
hc
hc
hc
hc
λ 31 =
=
= 300 nm
λ 41 =
=
= 200 nm
E3 − E1 4.14 eV
E4 − E1 6.21 eV
λ 21 =
hc
hc
=
= 500 nm
E2 − E1 2.49 eV
λ 42 =
hc
hc
=
= 334 nm
E4 − E2 3.72 eV
λ 43 =
hc
hc
=
= 601 nm
E4 − E3 2.07 eV
λ 32 =
hc
hc
=
= 753 nm
E3 − E2 1.65 eV
39.51. Model: Photons are emitted when an atom undergoes a quantum jump from a higher energy level to a
lower energy level. On the other hand, photons are absorbed in a quantum jump from a lower energy level to a
higher energy level. Because most of the atoms are in the n = 1 ground state, the only quantum jumps in the
absorption spectrum start from the n = 1 state.
Solve: (a) The ionization energy is E1 = 6.5 eV.
(b) The absorption spectrum consists of the transitions 1 → 2 and 1 → 3 from the ground state to excited states.
According to the Bohr model, the required photon frequency and wavelength are
f =
c hc
ΔE
⇒ λ= =
f ΔE
h
where ΔE = Ef – Ei is the energy change of the atom. Using the energies given in the figure, we calculated the
values in the table below.
Ei (eV)
Δ E (eV)
λ (nm)
Transition
Ef (eV)
1→2
1→3
−3.0
–2.0
−6.5
−6.5
3.5
4.5
355
276
(c) Both wavelengths are ultraviolet (λ < 400 nm).
(d) A photon with wavelength λ = 1240 nm has an energy Ephoton = hf = hc/λ = 1.0 eV. Because Ephoton must
exactly match Δ E of the atom, a 1240 nm photon can be emitted only in a 3 → 2 transition. So, after the collision
the
atom
was
in
the
n = 3 state. Before the collision, the atom was in its ground state (n = 1). Thus, an electron with vi = 1.4 × 106 m/s
collided with the atom in the n = 1 state. The atom gained 4.5 eV in the collision as it is was excited from the n =
1 to n = 3, so the electron lost 4.5 eV = 7.20 × 10−19 J of kinetic energy. Initially, the kinetic energy of the
electron was
K i = 12 melecvi2 =
1
2
( 9.11×10
kg )(1.40 × 106 m/s ) = 8.93 × 10 −19 J
2
−31
After losing 7.20 × 10−19 J in the collision, the kinetic energy is
K f = K i − 7.20 × 10−19 J = 1.73 × 10−19 J = 12 melecvf2 ⇒ vf =
2 (1.73 × 10−19 J )
2Kf
=
= 6.16 × 105 m/s
melec
9.11 × 10−31 kg
39.53. Solve: The radius of the orbit in state n is rn = n2aB. The quantum number is found as follows:
B
1
r
( 5.18 nm ) = 48.96 ≈ 49 ⇒ n = 7
n2 = n = 2
aB 0.0529 nm
Thus the energy is
En = −
13.60 eV
13.60 eV
⇒ E7 = −
= −0.278 eV
2
n
49
39.56. Solve: (a) From Equation 39.36, the wavelengths of the emission spectrum are
λ n→m =
91.18 nm
m −2 − n −2
m = 1, 2, 3, … n = m + 1, m + 2, …
For the 200 → 199 transition,
λ 200 →199 =
91.18 nm
(199 )
−2
− ( 200 )
−2
= 0.362 m
(b) For the 2 → 199 transition,
λ 2 →199 =
91.18 nm
( 2)
−2
− (199 )
−2
= 4.000404071 (91.18 nm)
Likewise, λ 2 → 200 = 4.00040004(91.18 nm). The difference in the wavelengths of these two transitions is
0.0000041 ×
(91.18 nm) = 0.000368 nm.
39.61. Solve: (a) Nearly all atoms spend nearly all of their time in the ground state (n = 1). To cause an emission
from
the
n = 3 state, the electrons must excite hydrogen atoms from the n = 1 state to the n = 3 state. The energy gained by each
atom is
E3 – E1 = −1.51 eV – (−13.60 eV) = 12.09 eV
This means the electrons must each lose 12.09 eV of kinetic energy. Thus the electrons must have at least Kmin =
12.09 eV of kinetic energy to cause the emission of 656 nm light. The minimum speed of the electrons is
vmin =
2 (12.09 eV ) 1.6 × 10−19 J
2 K min
=
×
= 2.06 × 106 m/s
melec
9.11 × 10−31 kg
1 eV
(b) An electron gains 12.09 eV of kinetic energy by being accelerated through a potential difference of 12.09 V.
This is simply the definition of electron volt.
39.63. Solve: The energy of the ultraviolet light photon is
E=
hc
λ
=
( 6.63 × 10
−34
J s )( 3.0 × 108 m/s )
70 × 10 −9 m
×
1 eV
= 17.76 eV
1.60 × 10−19 J
To ionize a hydrogen atom, a minimum energy of 13.60 eV is required. Thus, the kinetic energy of the freed
electrons is
K = 12 mv 2 = E − 13.60 eV = 17.76 eV − 13.60 eV = 4.16 eV