PH 300
Fall 2013
HW8 answer key
1) For an electron confined to a 2-dimensional box of length 0.1 nm,
what is the kinetic energy of the ground state? How does this
compare to the energy of an electron confined to a 1-dimensional
box?
From lecture notes: Enx,ny = -  2π2/2mL2(nx2 + ny2), ground state nx = 1, ny = 1
L = 0.1 nm, m = 0.511 MeV/c2, (  c)2 = (197.33 eV*nm)2
Enx,ny = -(  c)2π2/2mc2L2(nx2 + ny2)*2 = (37.6 eV)*2 = 75.2 eV
The one dimensional solution is En = n2  2π2/(2mL2) = 37.6n2 eV
2) An electron is trapped in an infinite square-well potential of width
0.5 nm. If the electron is initially in the n=4 state, what are the various
photon energies that can be emitted as the electron jumps to the
ground state? (T&R problem 6-21)
L = 0.5 nm, m = 0.511 MeV/c2, (  c)2 = (197.33 eV*nm)2
E1 = (  c)2π2/2mc2L2 = 1.504 eV, En = n2E1
E4-3 = (42 – 32)E1 = 10.53 eV
E4-2 = (42 – 22)E1 = 18.05 eV
E4-1 = (42 – 12)E1 = 22.56 eV
E3-2 = (32 – 22)E1 = 7.52 eV
E3-1 = (32 – 12)E1 = 12.03 eV
E2-1 = (22 – 12)E1 = 4.51 eV
3) How many different energy levels can a particle in a 2-dimensional
box have that are less than 60E0 where E0 = π2ħ2/2mL2? Show your
work.
(nx2 + ny2)E0 < 60E0
Note: No preferred direction ∴ energy levels are
symmetric for x,y values: E(3,1) = E(1,3) (degeneracy)
Start by finding max nx: nx = (60 -1)1/2 = (59)1/2 < 8 so nx = 7
E(7,1) → 50E0, E(7,2) → 53E0, E(7,3) → 58E0 this is the max
Now find the nx = ny max, 2n2 = 60 → n = (30)1/2 < 6 so n = 5
ny will never exceed 5 for nx → E(7,1) (Any that exceed 5 are repeats or
yield more than 60E0)
5 4 3 2 1
7 74 65 58 53 50 Color code:
6 61 52 45 40 37 *blue numbers are the possible energy levels
5 50 41 34 29 26 *black numbers represent combinations that do not
meet criteria
4 41 32 25 20 17
*red
numbers are degenerate solutions (repeats)
3 34 25 18 13 10
*green numbers are the combinations that are left
2 29 20 13 8 5
over
1 26 17 10 5 2
Note: E(7,1) and E(5,5) represent special degeneracy
So there are 21 allowable states (22 if counting
E(7,1) and E(5,5) as different energy states)
4) The harmonic oscillator potential is U(x)=(1/2)mω02x2; a particle of
mass m in this potential oscillates with frequency ω0. The groundstate wavefunction for a particle in the harmonic oscillator potential
has the form ψ(x)=Aexp(-ax2)
By substituting U(x) and ψ(x) into the one-dimensional timeindependent Schrodinger Equation, find expressions for the groundstate energy E and the constant a in terms of m, ħ, and ω0.
Start with the time independent Schrodinger Equation
−
 2 d 2ψ (x )
+ U ( x)ψ ( x) = Eψ ( x) w
2m d 2
x
, ψ ( x) = Ah
−a
2
a
1
x
eU
e( x)n= r mω02 xde2
2
To use the time independent Schrodinger Equation, a wavefunction must be
determined. So a normalization constant is needed.
∫
+∞
−∞
π
 2a 
=1→ A = 

2a
π 
x A2
A 2 e −2a d x = 1 →
2
1/ 4
Next take the first and second derivatives of the wave function
d
ψ ( x) = −2 x a −Aa
d x
d  d

ψ ( x)  = 4 A

x 
d  dx
2
x
e
2 −a
2
e a xx2 − 2a
−a
A
2
Sub parts into the time independent Schrodinger Equation and do some algebra
2

2  −  
1
2 ∗ (2a 2 x 2 − a )+ mω02 x 2 − E  = 0
Ae −ax 
2
 2m 

Then do some more algebra
 − 2 2 2 1
 
 2 a 
x 2 
a + mω 02  −  E −
=0
2
m 
 m
 
The quantities in parenthesis are both equal to zero, so
-2  2a2/m = -mω02/2 → a = mω0/2 
And
E =  2a/m = (  2/m)*(mω0/2 hbar) = (1/2)  ω0
5)The H2 molecule can be approximated by a simple harmonic
oscillator having spring constant k=1.1*10^3 N/m. Find (a) the energy
levels and (b) the possible wavelengths of photons emitted with the
H2 molecule decays from the third excited state eventually to the
ground state.
(a)
mm
m
and
=
m+m 2
the spring constant k=1100N/m. Thus the energies are En=(n+1/2)ħω, where
k
ω= 2
and m is the mass of H atom. And ħω=0.759ev=1.214*10^-19J
m
(b)
Ground state E0=1/2 ħω, and the third excited state E3=(3+1/2)ħω.
k
So ∆E=E3-E0=3 ħω=3 ħ 2 =3.644*10^-19J=2.278ev.
m
hc
= 545.8nm .
λ=
∆E
We simplify H2 as a harmonic oscillator, so the reduced mass is
6) Calculate the probability that an electron in the ground state of the
hydrogen atom can be found between 0.95a0 and 1.05a0.
The ground state for the hydrogen atom is n = 0, l = 0, ms = 0
Ψ(r) →
R nl (r ) =
2
a 3/2
0
e -r/a 0
And

2  2
Pnl (r ) = r ⋅ 3 / 2
a
 0
2
 − 2 r/a
0
 e


So the probability is
1.05a 0
4
a 30
∫
r 2 e − 2 r/a 0 dr = 5.41%
0.95a 0
7) The 3-D Schrödinger Equation (S.E.) for the hydrogen atom looks
like:
∂ 2ψ 
∂ψ 
1
1 1 ∂ 
 2  ∂ 2ψ 2 ∂ψ
+
+
−
+
 sin θ
 + U (r )ψ = Eψ

∂θ  r 2 sin 2 θ ∂φ 2 
2m  ∂r 2 r ∂r r 2 sin θ ∂θ 
where
show that ψ (r , θ , φ ) =
E=−
U (r ) = −
1
π a0 3 / 2
1
e
4πε 0 r
e − r / a0 satisfies this equation provided
me 4
32π 2 ε 02  2 (ground state energy of the H atom)
and a0 (Bohr radius)
∂ 2ψ + e − r / a0
∂ψ − e − r / a0
∂ψ ∂ψ
=
=
=
=0
Take derivatives: (a)
,
(b)
,
(c)
∂r
∂θ
∂φ
∂r 2
π a07 / 2
π a03 / 2
plug (a) (b) (c) into the time independent S.E.
(-  2/2m)*[∂2ψ/∂r2 + 2/r*∂ψ/∂r + 0 + 0] + U(r) ψ(r) = E*ψ (r)
 − 2 e− r / a0   − 2 2 e− r / a0   e2
e − r / a0 
e− r / a0

 
 

 2m ⋅ π a 7 / 2  +  2mr ⋅ π a 5 / 2  −  4πε r ⋅ π a 3 / 2  = E ⋅ π a 3 / 2
0
0
0
0
0

 
 

 − 2 1   − 2 2 1   e2 
 = E

⋅  − 
⋅ 2  + 
m
a
mr
a
r
2
2
4
πε
0 
0 
0 



 − 2 1 
1  −  2 2 1   e 2  

 = 0
⋅ 2  − E + 
⋅  − 
m
a
r
m
a
πε
2
2
4
0 
0 
0 



For this to be true for all r,
 − 2 2 1

⋅
 2m a 0
  e2 
 = 

  4πε 0 
(solve for a0) a0 = (4πε0  2)/me2
Also…
 2 1 

⋅ 2  = E
 2m a 0 
Plug in a0 to find E using above equation to get
E = 1/a0*e /8πε0 and E = −
2
me 4
32π 2 ε 02  2
Extra Credit:
E=65E0, 85E0, 125E0
Need to attach you calculations or your computer program.