Problems and Questions on Lecture 4
1. Physical correspondence of simple harmonic oscillator potential is
(A) vibration of diatomic molecules (B) free electron, (C) hydrogen atom, (D) freely falling bodies (E) rotation of earth
2. Physical correspondence of Coulomb potential is
(A) vibration of diatomic molecules (B) free electron, (C) hydrogen atom, (D) freely falling bodies (E) rotation of earth
3. Physical correspondence of zero potential is
(A) vibration of diatomic molecules (B) free electron, (C) hydrogen atom, (D) freely falling bodies (E) rotation of earth
4. Harmonic oscillator potential is given by
𝑍𝑒 2
1
1
(A) 𝑉 =
; (B) 𝑉 = 𝑚𝜔2 𝑥 2 ; (C) 𝑉 = − 𝑚𝜔2 𝑥 2 ; (D) 𝑉 = −
𝑟
2
2
5. Coulomb potential for an H-like atom is given by
𝑍𝑒 2
1
1
(A) 𝑉 =
; (B) 𝑉 = 𝑚𝜔2 𝑥 2 ; (C) 𝑉 = − 𝑚𝜔2 𝑥 2 ; (D) 𝑉 = −
𝑟
2
2
6. Graph of the harmonic oscillator potential is
8
4
2
2
6
2
4
4
2
6
𝑍𝑒 2
𝑟
𝑍𝑒 2
𝑟
; (E) 𝑉 = 0
; (E) 𝑉 = 0
4
1
2
3
4
1
2
3
4
0.5
1.0
1.5
2.0
2.5
(A)
4
2
2
4
(B)
(C)
8
3.0
3.0
2.5
𝑉
𝑉 =∞
=0
0
𝑎
𝑥
2.0
1.5
1.0
0.5
(D)
1
2
3
4
(E)
7. Graph of the Coulomb potential is
8
4
2
2
4
0.5
2
6
1.0
4
4
2
6
1.5
2.0
2.5
(A)
4
2
2
4
(B)
(C)
8
3.0
3.0
2.5
𝑉
𝑉 =∞
=0
(D)
(E)
8. A particle in the infinite well potential
0 of𝑎width𝑥[0,a]. Its quantum number 𝑛 and the integral is
2.0
1.5
1.0
0.5
1
2
3
4
2
2
𝑛𝜋
𝑥
1
2𝑛𝜋
∫ (√ sin ( 𝑥)) 𝑑𝑥 = −
sin (
𝑥).
𝑎
𝑎
𝑎 2𝑛𝜋
𝑎
Probability of finding of particle between [0,a] is
1
2
(A) 1, (B) 0.5, (C) 0.25
(D) (1 − )
(E) cannot be determined
4
𝑛𝜋
9. A particle in the infinite well potential of width [0,a/2]. Its quantum number 𝑛 and the integral is
2
2
𝑛𝜋
𝑥
1
2𝑛𝜋
∫ (√ sin ( 𝑥)) 𝑑𝑥 = −
sin (
𝑥).
𝑎
𝑎
𝑎 2𝑛𝜋
𝑎
Probability of finding of particle between [0,a] is
1
2
(A) 1, (B) 0.5, (C) 0.25
(D) (1 − )
(E) cannot be determined
4
𝑛𝜋
10. A particle in the infinite square well has its initial wave function is a mixture of first two stationary states:
Ψ(𝑥, 0) = 𝐴(𝜓1 (𝑥) + 𝜓2 (𝑥)).
Normalization constant A is
2
(C) √ ;
(A) 1, (B) √2,
𝑎
(D) 1/√2;
(E) 2
11. A particle in the infinite square well has its initial wave function is a mixture of first two stationary states:
Ψ(𝑥, 0) = 𝐴(𝜓1 (𝑥) + 𝜓2 (𝑥)).
Time dependent wave function Ψ(𝑥, 𝑡) is
𝐸1
𝐸2
𝐸1
𝐸2
(A) 𝐴 (𝜓1 (𝑥)𝑒 −𝑖 ℏ 𝑡 + 𝜓2 (𝑥)𝑒 −𝑖 ℏ 𝑡 ) ; (B) 𝐴 (𝜓1 (𝑥)𝑒 − ℏ 𝑡 + 𝜓2 (𝑥)𝑒 − ℏ 𝑡 ),
𝑖𝐸𝑛
𝑡
ℏ
(C) 𝐴𝑒 −
(𝜓1 (𝑥) + 𝜓2 (𝑥));
𝐴(𝜓1 (𝑥) + 𝜓2 (𝑥)) (E) none of them
12. A particle in the infinite square well has its initial wave function is a mixture of first two stationary states:
Ψ(𝑥, 0) = 𝐴(𝜓1 (𝑥) + 𝜓2 (𝑥)).
2
|Ψ(𝑥,
Norm of wave function
𝑡)| is
(D)
3ℏ𝜋2 𝑡
(A) 1, (B) 𝐴2 (|𝜓1 |2 + |𝜓1 |2 + 𝜓1 𝜓2 2cos [
]),
2𝑚𝑎2
2 (|𝜓 |2
2)
|𝜓1 |2 + 𝜓1 𝜓2 );
|𝜓
|
(E) 𝐴
1 +
1
13. Schrödinger equation of a particle on a ring is
(A) −
ℎ
𝑑
2𝑚 𝑑𝑥
𝜓 = 𝐸𝜓
(B) −
ℏ2
𝜕2 𝜓
2𝑚𝑅 2 𝜕𝜙2
= 𝐸𝜓
(C) 𝐴2 (|𝜓1 |2 + |𝜓1 |2 + 𝜓1 𝜓2 2sin [
(C) 𝜋𝑟 2 𝜓 = 𝐸𝜓
3ℏ𝜋2 𝑡
2𝑚𝑎2
(D) 𝜓 = 𝑐𝑜𝑛𝑠 𝑒 ∓𝑖𝑚𝜙
]);
(D) 𝐴2 (|𝜓1 |2 +
(E) 𝐸 =
ℏ2
2𝐼
𝑚2
14. Energy of the particle on a ring is
ℏ2
13.6
𝑛2 ℏ2 𝜋2
1
(A) 𝐸 = 𝑚2 ;
(B) 𝐸 = 𝑛ℎ𝜈;
(C) 𝐸 = − 2 ;
(D) 𝐸 = (𝑛 + ) ℏ𝜔
(E) 𝐸 =
2𝐼
𝑛
2
2𝑚𝑎2
15. Uncertainty between 𝐿𝑧 𝑎𝑛𝑑 𝜙 is
ℏ
ℏ
(A) 0 (B) ℏ
(C)
(D) 𝑖ℏ
(E) −
2
2
16. Radius of the benzene molecule is 1.39Å. Free electron of the molecule is on a ring and make transition from 1st level to 2nd
level. The wavelength of emitted photon is
(A) in the x-ray region, (B) in the visible region (C) in the UV region, (D) in the far infrared region
(E) in the gamma
ray region
17. What is the SE of a particle in the 3dimensional box?
18. What is the energy of a particle in 3 dimensional box?
19. Calculate normalization constant for the wave function of a particle in 3D box.
𝜋2 ℏ2
20. The ground state energy of the 2D box of size 𝐿 𝑥 𝐿 is 2𝐸0 where 𝐸0 =
is the ground state energy of a 1D box of size L.
2𝑚𝐿2
First excited state of this 2D box is
(A) 2𝐸0
(B) 3𝐸0
(C) 4𝐸0
(D) 5𝐸0 (E) 3𝐸0
21. If energy of a particle in the box is represented by 𝐸𝑛𝑥,𝑛𝑦 ,𝑛𝑧 , where 𝑛𝑥 , 𝑛𝑦 , 𝑎𝑛𝑑 𝑛𝑧 are quantum numbers. Ground state energy
of the particle in 3D box is
(A) 𝐸111 , (B) 𝐸001
(C) 𝐸100
(D) 𝐸123 (E) 𝐸000
22. An particle in a 1D rigid box of length L (infinite square well) is in it ground state. The box is compressed to length L/2. (a)
By what factor does the ground-state energy increase? (b) By what factor does the first excited state energy increase? (c)
Repeat the calculations in (a) and (b) for the case of a particle in a 3D box of volume V = L 3.
23. The stars does not collapse because of
(A) degenerate pressure (B) gravity (C) electric pressure (D) magnetic pressure (E) energy
24. What is the Fermi energy?
25. Find Fermi energy of a particle in the box?
26. Please take attention the the examples solved in the class!!!!
27. What is the importance of triangular potential well?
28. Find degeneracy pressure on the walls of particle in the cubic box
29. Please study derivation of wave function and energy of a particle in the square box.
30. Find transmission and reflection coefficients of matter from step potential and barrier potential.
31. The wave function for a particle confined in an infinite square well arranged to the coordinate system [0, 𝑎] is given by:
𝑖𝐸
𝑛𝜋
𝛹(𝑥, 𝑡) = 𝑒 − ℏ 𝑡 (𝐴𝑠𝑖𝑛[ 𝑥])
𝑎
where 𝑛 = 1,2,3, …, E corresponds to energy, m is mass of the particle, A is constants. ℏ is Planck's constant.
𝑎
𝑛𝜋
𝑛𝜋
𝑎
a) Calculate A. (Hint: use normalization and ∫0 sin ( 𝑥) sin ( 𝑥) 𝑑𝑥 = )
𝑎
𝑎
2
b) Is 𝛹(𝑥, 𝑡) valid wave function? Explain.
𝜕Ψ(𝑥,𝑡)
ℏ2 𝜕2 Ψ(𝑥,𝑡)
Verify that Ψ(𝑥, 𝑡) satisfies Schrödinger’s equation 𝑖ℏ
=−
., for some certain values of energy E. (Note
𝜕𝑡
2𝑚 𝜕𝑥 2
that for the region inside the well 𝑉(𝑥) = 0.) What is the values of E?
d) Calculate uncertainty between x and p, by using variance of x and p.
𝑎
𝑎
e) What is the probability of finding particle beween [0, ], [0, ].
2
4
Hint:
𝑎
𝑛𝜋
𝑛𝜋
𝑎2
∫ sin (
𝑥) (𝑥) sin (
𝑥) 𝑑𝑥 =
𝑎
𝑎
4
0
3
𝑎
𝑛𝜋
𝑛𝜋
𝑎
3
∫ sin (
𝑥) (𝑥 2 ) sin (
𝑥) 𝑑𝑥 =
(2 − 2 2 )
𝑎
𝑎
12
𝑛 𝜋
0
c)
𝑎
𝑛𝜋
𝜕
𝑛𝜋
𝑥) (−𝑖ℏ ) sin (
𝑥) 𝑑𝑥 = 0
𝑎
𝜕𝑥
𝑎
0
𝑎
𝑛𝜋
𝜕 2
𝑛𝜋
ℏ2 𝑛 2 𝜋 2
∫ sin (
𝑥) (−𝑖ℏ ) sin (
𝑥) 𝑑𝑥 =
𝑎
𝜕𝑥
𝑎
2𝑎
0
2𝑏𝑛𝜋
𝑏
𝑛𝜋
𝑛𝜋
𝑏 𝑎Sin[ 𝑎 ]
∫ sin (
𝑥) sin (
𝑥) 𝑑𝑥 = −
𝑎
𝑎
2
4𝑛𝜋
0
32. (Griffiths 1.4) At time t=0 a particle is represented by the wave function:
∫ sin (
𝑥
𝑎
𝑏−𝑥
Ψ(𝑥, 0) =
𝐴
𝑏−𝑎
{ 0
𝐴
𝑖𝑓 0 ≤ 𝑥 ≤ 𝑎
𝑖𝑓 𝑎 ≤ 𝑥 ≤ 𝑏
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
where A, a and b are constants.
a) Normalize Ψ (That is, find A, in terms of a and b).
b) Sketch Ψ(x, 0) as a function of x.
c) Where is the particle most likely to be found at t=0?
d)What is the probability of finding the particle to left of a? Check your result in the limiting cases 𝑏 = 𝑎 and 𝑏 = 2𝑎.
e) What is the expectation value of 𝑥?
Solution 15.
a) Using normalization condition:
∞
∫ Ψ ∗ Ψ𝑑𝑥 = 1
−∞
𝑎
1 = |𝐴|2 ∫
0
2
𝑥
(𝑏 − 𝑥)2
𝑎 𝑏−𝑎
3
2
|𝐴|
𝑑𝑥
+
∫
𝑑𝑥 = |𝐴|2 ( +
) => 𝐴 = √
2
𝑎2
(𝑏
−
𝑎)
3
3
𝑏
0
𝑎
b) Graph of the function:
c) To obtain most probable point sketch graph Ψ∗ Ψ then you the maximum point 𝑥 = 𝑎 is also most probable point.
𝑎 𝑥2
d) 𝑃 = |𝐴|2 ∫0
𝑎2
𝑑𝑥 =
3𝑎
𝑏3
𝑎
= .
𝑏
∞
e) Expectation value of an operator: ⟨𝑂⟩ = ∫−∞ Ψ∗ 𝑂Ψ𝑑𝑥
𝑎 3
𝑎 (𝑏
𝑥
− 𝑥)𝑥(𝑏 − 𝑥)
2𝑎 + 𝑏
⟨𝑥⟩ = |𝐴|2 ∫ 2 𝑑𝑥 + |𝐴|2 ∫
𝑑𝑥 =
2
𝑎
(𝑏
−
𝑎)
4
0
0
33. (Griffiths Problem 1.17). A particle is represented (at time t=0) by the wave function
𝐴(𝑎2 − 𝑥 2 ) 𝑖𝑓 − 𝑎 ≤ 𝑥 ≤ +𝑎
Ψ(𝑥, 0) = {
0
𝑂𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
a) Determine the normalization constant A.
b) What is the expectation value of x?
c) What is the expectation value of momentum p?
d) Find the uncertainty in x.
e) Find the uncertainty in p.
f) Check your results are consistent with the uncertainty principle.
Solution 16
a) Again using the relation
∞
∫ Ψ∗ Ψ𝑑𝑥 = 1
−∞
𝑎
2 1
15
1 = |𝐴|2 ∫ (𝑎2 − 𝑥 2 )2 𝑑𝑥 = 𝑎|𝐴|2 𝑎5 (1 − + ) => 𝐴 = √
3 5
16𝑎5
−𝑎
b) Expectation value of x:
⟨𝑥⟩ =
c) Expectation value of p:
⟨𝑝⟩ =
15 𝑎 2
∫ (𝑎 − 𝑥 2 )𝑥(𝑎2 − 𝑥 2 )𝑑𝑥 = 0
16𝑎5 −𝑎
15 𝑎 2
𝑑
∫ (𝑎 − 𝑥 2 ) (−𝑖ℏ ) (𝑎2 − 𝑥 2 )𝑑𝑥 = 0
16𝑎5 −𝑎
𝑑𝑥
d) Uncertainty in x:
⟨𝑥 2 ⟩ =
15 𝑎 2
𝑎2
2 )𝑥 2 2
2
(𝑎
∫
−
𝑥
(𝑎
−
𝑥
)𝑑𝑥
=
16𝑎5 −𝑎
7
e) Uncertainty in p:
⟨𝑝2 ⟩ =
f) Uncertainty relation:
15 𝑎 2
𝑑 2 2
5 ℏ2
2)
2
(𝑎
∫
−
𝑥
(−𝑖ℏ
)
(𝑎
−
𝑥
)𝑑𝑥
=
16𝑎5 −𝑎
𝑑𝑥
2 𝑎2
Δ𝑥Δ𝑝 = √⟨𝑥 2 ⟩ − ⟨𝑥⟩2 √⟨𝑝2 ⟩ − ⟨𝑝⟩2 = √
5
ℏ
14
𝑝2
𝑑
34. Under what condition the function 𝜓(𝑥) = 𝐴𝑐𝑜𝑠𝑘𝑥 is eigenfunction of the operator 𝐻 =
+ 𝑉0 where 𝑝 = −𝑖ℏ
is
2𝑚
𝑑𝑥
momentum operator, if its eigenvalue is twice of the constant potential 𝑉0 .
Solution
𝐻𝜓 = 𝐸𝜓
ℏ2 𝑑 2
−
𝐴𝑐𝑜𝑠𝑘𝑥 + 𝑉0 𝐴𝐶𝑜𝑠𝑘𝑥 = 2𝑉0 𝐴𝑐𝑜𝑠𝑘𝑥
2𝑚 𝑑𝑥 2
2
ℏ 2
2𝑚
𝑘 + 𝑉0 = 2𝑉0 𝑜𝑟 𝑘 2 = 2 𝑉0
2𝑚
ℏ
35. Why do we say the wavefunction completely specifies the state of a system? How do we use the wavefunction?
36. The nuclear potential energy can be modelled as a barrier potential as in the figure. Energy of alpha particles
within the nucleus less than height of the nuclear potential barrier, but has some chance tunneling through the barrier.
For values of = 1, V0 = 1 (𝑢𝑛𝑖𝑡𝑠) , transmission probability is given in the table.
Barrier with
height 𝑉0
inside
nucleus
0
outside nucleus
L
x
Energy
(Unit)
0.
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Transmission #particle pass
Probability
Through barrier
0.
0.23
0.38
0.49
0.57
0.63
0.68
0.72
0.75
0.78
a) If 1000 particle incident on the barrier, calculate number of particle pass through the barrier.
b) Plot graph of transmission probability versus energy on the graph region.
37. Wave function of a particle in a potential well shown in figure is given by
𝜓1 =
𝜓2 =
2𝑘 −𝑖𝑘′𝑥
𝑒
𝑘 + 𝑘′
𝑘 − 𝑘 ′ 𝑖𝑘𝑥
𝑒 + 𝑒 −𝑖𝑘𝑥
𝑘 + 𝑘′
𝜓3 = 𝐹𝑒 𝑖𝑘′𝑥
a) Calculate F.
Finite Potential Well
b) Show that the following relations are satisfied.
2
4
2
2𝑖𝑘𝑘′
𝑖(𝑘 + 𝑘′ )
= 𝑡𝑎𝑛[𝑘𝐿] 𝑜𝑟
= 𝑐𝑜𝑡[𝑘𝐿]
2
+ 𝑘′
2𝑘𝑘′
3
𝑘2
2𝑚𝐸
;
ℏ2
where 𝑘 = √
2
2𝑚
𝑘′ = √ ℏ2 (𝐸 − 𝑉0 )
1
0
1
38. Expectation values of energy of a particle in the
2
2
4
6
x
infinite well box of width a is given by
⟨𝐸⟩ =
0
ℏ2 𝑛 2 𝜋 2
2𝑚𝑎2
Where 𝑛 is quantum number 𝑛 = 1, 2, 3 ….
a) If width of the well is 𝑎 = 1Å, what is the minimum energy of a electron in the box in the unit of eV.
(𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛, 𝑚 = 9.11 × 10−31 𝑘𝑔; 1Å. , = 10−10 𝑚, 𝑒 = 1.6 × 10−19 𝐶 )(Just estimate order of energy in eV).
b) Estimate energy of an electron in infinite well of width 𝑎 = 1Å, using uncertainty relation.
c) Calculate kinetic energy of an electron of wavelength 𝜆 = 1Å, using de Broglie relation and compare it to energy of a
particle in part (a and b).( Again estimate order of energy in eV)
d) Repeat part (a), and (b) for a proton in the well of width 10−15 𝑚. Compare your result to energy of electron in part (a)
and (b). (mass of proton, 𝑚 = 1.67 × 10−27 𝑘𝑔.)