I. In Problem 3, the energy E of particle in one- dimensional box is calculated as: 2 2 2= n 8 L h E — Analytical Chemistry Chemistry Question
Particles in 2, 3 - Dimensional Box
I. In Problem 3, the energy E of particle in one- dimensional box is calculated as:
2 2
2= n 8 L
h E
m
where h is Planck’s constant; m is the mass of the particle; L is the box length;
n is the quantum number, n = 1, 2, 3…
An electron in a 10 nm one-dimensional box is excited from the ground state to a higher energy level by absorbing a photon of the electromagnetic radiation with a wavelength of 1.374 ⋅ 10-5 m.
What is the energy gap (∆E) of the two mentioned states?
Model Answer
4.1
2 2 2 2 2 2 2
2 2 2∆ = n 1 = (n 1 ) 8 L 8 L 8 L
h h h E
m m m – –
Determine the final energy state for this transition.
Model Answer
4.2 According to Planck’s equation:
-34 8 -1
-20 -5
(6.626 10 J s)(2.9979 10 m s ) ∆ = = = 1.446 10 J
1.374 10 m h c
E λ
⋅ ⋅ ⋅ ⋅
( ) -34 2
-20 2 -31 -9 2
(6.626 10 J s) ∆ = 1.446 10 J = n 1
8(9.109 10 kg)(10.0 10 m) E
⋅⋅ ⋅ ⋅
–
( )-20 -22 2
2 2
1.446 10 J = 6.025 10 n 1
n 1= 24.00 n = 25.00
⋅ ⋅
→ → →
–
– n = 5.00
II. The treatment of a particle in a one- dimensional box can be extended to a two- dimensional box of dimensions Lx and Ly yielding the following expression for energy:
222 yx
2 2 x y
nn = +
8 L L h
E m
The two quantum numbers independently can assume only integer values. Consider an electron confined in a two-dimensional box that is Lx = 8.00 nm in the x direction and Ly = 5.00 nm in the y direction.
4.3 What are the quantum numbers for the first three allowed energy levels? Write the first three energy, Exy, in order of increasing energy?
Model Answer
4.3 The quantum numbers are:
Ground state (E11) → nx = 1, ny = 1
First excited state (E21) → nx = 2, ny = 1
Second excited state (E12) → nx = 1, ny = 2
Since the energy levels, Exy, are inversely proportional to L2, then the nx = 2, ny = 1 energy level will be lower than the nx = 1, ny = 2 energy level since Lx > Ly.
The first three energy levels, Exy, in order of increasing energy are: E11 < E21 < E12
Calculate the wavelength of light necessary to move an electron from the first excited state to the second excited one.
Model Answer
4.4 Calculate the wavelength of light necessary to promote an electron from the first excited state to the second excited state.
E21 → E12 is transition. 222 yx
xy 2 2 x y
nn = +
8 L L h
E m
2 2 2 17 2
12 -9 2 -9 2
2 2 2 17 2
21 -9 2 -9 2
17 2 17 2 16 2
12 21
16 -2
1 2 1.76 10 = + =
8 (8.00 10 m) (5.00 10 m) 8
2 1 1.03 10 = + =
8 (8.00 10 m) (5.00 10 m) 8
1.76 10 1.03 10 7.3 10 ∆ = - = =
8 m 8 m 8 m
(7.3 10 m )(6.626 10 ∆ =
h h E
m m
h h E
m m
h h h E E E
E
⋅ ⋅ ⋅
⋅ ⋅ ⋅
⋅ ⋅ ⋅−
⋅ ⋅ -34 2 -21
-31
-34 8 -1 -5
-21
Js) = 4.4 10 J
8 (9.11 10 kg)
(6.626 10 J s) (2.998 10 m s ) = = = 4.5 10 m ∆ 4.4 10 J h c
E λ
⋅ ⋅
⋅ ⋅ ⋅ ⋅
III. Similarly, the treatment of a particle in a one-dimensional box can be extended to a rectangular box of dimensions Lx, Ly, and Lz, yielding the following expression for energy:
22 22 yx z
2 2 2 x y z
nn n = + +
8 L L L h
E m
The three quantum numbers nx, ny, and nz independently can assume only integer values.
An oxygen molecule is confined in a cubic box of volume 8.00 m3. Assume that the molecule has an energy of 6.173 ⋅ 10–21 J; temperature T = 298 K.
4.5 What is the value of 2 2 2 1/2 x y z= ( + + )n n n n for this molecule?
Model Answer
4.5 2 2 2 2 2 2
-211 2 3 2 2
(n +n +n ) n = = = 6.173 10 J
8 8 h h
E m L m L
⋅
2 2
2
8 L n =
m E
h
If L3 = 8.00 m3, then L2 = 4.00 m2
2 -34 2 -43
2
23
(6.626 10 ) = = 2.582 10 J
8 L 0.032 8 4
6.022 10
h m
⋅ ⋅
⋅
-21 2 22 11
-43
6.173 10 n = = 2.39 10 ; n =1.55 10
2.582 10 ⋅ ⋅ ⋅ ⋅
In a rough approximation to estimate the energy gap of this system, assume that the two closest energy levels correspond to n and n + 1.
4.6 What is the energy separation between the levels n and n + 1?
Model Answer
4.6 11 11n+1 n 1.55 10 +1 1.55 1× 0× ∆ = =E E E E E– –
2 2
11 -31 2 2∆ = (2n+1) = [2(1.55 10 +1] = 8.00 10 J
8 L 8 )
L h h
E m m
⋅ ⋅
IV. In quantum mechanics, an energy level is said to be degenerate if it corresponds to two or more different measurable states of a quantum system. Consider a particle in a cubic box.
4.7 What is the degeneracy of the level that has energy 21/3 times that of the lowest level?
Model Answer
4.7 The energy levels are
1 2 3
2 2 2 2 2 2 21 2 3
n ,n ,n 1 1 2 32
(n +n +n )h = = (n +n +n )
8 L E E
m
where E1 combines all constants besides quantum numbers. The minimum value for all quantum numbers is 1, so the lowest energy is
E1,1,1 = 3E1
The question asks about an energy 21/3 times this amount, namely 21E1. This energy level can be obtained by any combination of allowed quantum numbers such that 2 2 2 1 2 3(n +n +n )= 21 = 42 + 22 + 12
The degeneracy, then is 6, corresponding to (n1, n2, n3) = (1, 2, 4), (1, 4, 2), (2, 1, 4), (2, 4, 1), (4, 1, 2), or (4, 2, 1).