1237.Quantum Mechanics 1
Transcript of 1237.Quantum Mechanics 1
The Birth of Quantum MechanicsAt the turn of the last century, there were several experimental observations which could not be explained by the established laws of classical physics and called for a radically different way of thinking.
This led to the development of Quantum Mechanics which is today regarded as the fundamental theory of Nature.
NEED:Some key events/observations that led to the development of quantum mechanics… Black body radiation spectrum (Planck, 1901)
Photoelectric effect (Einstein, 1905)
Model of the atom (Bohr’s 1926)
Quantum Theory of Spectra (Bohr, 1913)
Scattering of photons off electrons (Compton, 1922) Matter Waves (de Broglie 1925)
Model of atom
Rutherford model
Classical models
Bohr model of atom-Quantum model
Quantum theory of spectra (Origin of Characteristic x-rays )
hv
Hydrogen spectra
Black Body Radiation
A black body is an object that absorbs 100% of the radiation that hits it. Therefore it reflects no radiation and appears perfectly black.
Wien’s Displacement Law:
Stefan’s Law (1879): the total power P radiated from one square meter of black surface at temperature T goes as the fourth power of the absolute temperature:
4 8 4, 5.67 10 watts/sq.m./K .P T
KoA7102.898constant.Tmλ
Blackbody Approximation
A good approximation of a black body is a small hole leading to the inside of a hollow object
The hole acts as a perfect absorber
The nature of the radiation leaving the cavity through the hole depends only on the temperature of the cavity
Rayleigh-Jeans Law:
An early classical attempt to explain blackbody radiation was the Rayleigh-Jeans law
At long wavelengths, the law matched experimental results fairly well
4
2I , Bπck T
λ Tλ
Rayleigh-Jeans Law, cont.
At short wavelengths, there was a major disagreement between the Rayleigh-Jeans law and experiment
This mismatch became known as the ultraviolet catastrophe
Planck’s Quantum Theory
In 1900 Planck devised a theory of blackbody radiation which gave good agreement for all wavelengths. In this theory the molecules of a body cannot have arbitrary energies but instead are quantized - the energies can only have discrete values. The magnitude of these energies is given by the formula
E = nhv,
where n = 0,1,2,... is an integer, v is the frequency of vibration of the molecule, and h is a constant, now called Planck's constant:
h = 6.63 x 10- 34 Js .
Classical predictions of black body radiation
hv2hv
Quantum prediction
Planck’s Model, Graphs
Planck’s Wavelength Distribution Function Planck’s Wavelength Distribution Function Planck generated a theoretical expression for the
wavelength distribution
h = 6.626 x 10-34 J.s h is a fundamental constant of nature
2
5
2
1I ,
Bhc λk T
πhcλ T
λ e
The photoelectric effect
Another experiment which provides compelling proof for the photon nature of light is the photoelectric effect.
hvh-W0=1/2mv2
Classical view Quantum view
Photoelectric Effect Summary
Each metal has “Work Function” (W0) which is the minimum energy needed to free electron from atom.
Light comes in packets called Photons E = h v h=6.626 X 10-34 Joule sec
Maximum kinetic energy of released electrons K.E. = hv – W0
PHOTOELECTRIC EFFECT (cont)
The maximum KE of an emitted electron is then
maxK h W
Work function: minimum energy needed for electron to escape from metal (depends on material, but usually 2-5eV)
Planck constant: universal constant of nature
346.63 10 Jsh
Einstein
Millikan
Verified in detail through subsequent experiments by Millikan
Maximum KE of ejected electrons is independent of intensity, but dependent on ν
For ν<ν0 (i.e. for frequencies below a cut-off frequency) no electrons are emitted
There is no time lag. However, rate of ejection of electrons depends on light intensity.
Actual results:
E h
Einstein’s interpretation (1905):
Light comes in packets of energy (photons)
An electron absorbs a single photon to leave the material
h (- 0) = 1/2mv2
Numerical Questions related to photo electric effect
Q1. Photoelectric work function of a photo-sensitive material is 3 eV. Calculate its threshold frequency and threshold wavelength.
Hint:
h (- 0) = 1/2mv2
h - h0 = 1/2mv2
As h0 = W = 3 eV
hvW
Therefore 0 = W/h = 3 x 1.6 x 10 -19 /6.62 10 -34 Hz
And 0 = c/ 0 Ao
1/2mv2
Numerical Questions related to photo electric effect
Q 1.5 mW of 400 nm light is directed at a photoelectric cell. If 0.10 percent of the incident photons produce photoelectrons, find the current in the cell.
Hint:
.483.0
10483.0
106.1sec1002.3
.sec1002.3.100
10302.010.0
10.0
.1030212.0103.3
109375.0
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Numerical Questions related to photo electric effect
Q Light from the sun arrives at the earth , an average of 1.5 x 1011 m away, at the rate of 1.4 x 103 W/m2 of area perpendicular to the direction of light. Assume that sun light is monochromatic with a frequency of 5.0 x 1014 Hz, how many photons fall per second on each square meter of the earth surface directly facing the sun?
Hint:
.sec/102.4100422.0100.51062.6
104.1 21231434
3
photonsn
PtnhE
COMPTON SCATTERING (cont)
Compton’s explanation:“billiard ball” collisions between particles of light (X-ray photons) and electrons in the material
Classical picture:oscillating electromagnetic field causes oscillations in positions of charged particles, which re-radiate in all directions at same frequency and wavelengthas incident radiation.
Change in wavelength of scattered light is completely unexpectedclassically
θ
ep
pBefore After
Electron
Incoming photon
p
scattered photon
scattered electron
Oscillating electronIncident light wave Emitted light wave
Experiment: Outgoing photon has longer wavelength
Recoil electron carries some momentum and KE
Incoming photon has momentum, p, and wavelength
This experiment really shows photon momentum!
Electron at rest
Compton Scattering
Pincoming photon + 0 = Poutgoing photon + Pelectron
hc
hvE h
p
Energy of a photon
Momentum of a photon
Conservation of energy Conservation of momentum
1/ 22 2 2 2 4e e eh m c h p c m c ˆ
e
h p i p p
1 cos
1 cos 0e
c
h
m c
12 Compton wavelength 2.4 10 mce
h
m c
From this Compton derived the change in wavelength
θ
ep
pBefore After
Electron
Incoming photon
p
scattered photon
scattered electron
COMPTON SCATTERING (cont)
Numerical Questions related to Compton effect
Q1 An x-ray photon of frequency 3 x 10 19 Hz collides with an electron and is scattered through an angle of 900. Find its new frequency.
Hint:
1 cos
1 cos 0e
c
h
m c
The nature of light…….
The birth of quantum mechanics is intimately linked with the theories and discoveries relating to the nature of light
Is the nature of light that of a wave or a particle???
Light has a dual nature
Wave (electromagnetic) - Interference - Diffraction Particle (photons) - Photoelectric effect - Compton effect
Wave - Particle Duality for light
The connecting link – Planck’s constant_______________________________
Dual Nature
Radiation
Matter
hE
p
h
SUMMARY OF PHOTON PROPERTIESSUMMARY OF PHOTON PROPERTIES
E h
h hp
c
E p k2
h
2k
Energy and frequency
Also have relation between momentum and wavelength
2 2 2 2 4E p c m c
c
Relation between particle and wave properties of light
Relativistic formula relatingenergy and momentum
E pcFor light and
Also commonly write these as
2 angular frequency
wavevector
hbar
WAVE-PARTICLE DUALITY OF LIGHTIn 1924 Einstein wrote:- “ There are therefore now two theories of light, both indispensable, and … without any logical connection.”
Evidence for wave-nature of light• Diffraction and interferenceEvidence for particle-nature of light• Photoelectric effect• Compton effect
•Light exhibits diffraction and interference phenomena that are only explicable in terms of wave properties
•Light is always detected as packets (photons); if we look, we never observe half a photon
•Number of photons proportional to energy density (i.e. to square of electromagnetic field strength)
We have seen that light comes in discrete units (photons) with particle properties (energy and momentum) that are related to thewave-like properties of frequency and wavelength.
MATTER WAVES
h
p
In 1923 Prince Louis de Broglie postulated that ordinary matter can havewave-like properties, with the wavelength λ related to momentum p in the same way as for light
de Broglie wavelength
de Broglie relation346.63 10 Jsh
Planck’s constant
Prediction: We should see diffraction and interference of matter waves
De Broglie
NB wavelength depends on momentum, not on the physical size of the particle
ACT: De Broglie WavelengthA stone is dropped from the top of a building.
1. It decreases
2. It stays the same
3. It increases
What happens to the de Broglie wavelength of the stone as it falls?
p
h
hp
Speed, v, KE=mv2/2, and momentum, p=mv, increase.
Activity:Photon A has twice as much momentum as Photon B. Compare their energies.
• EA = EB
• EA = 2 EB
• EA = 4 EB
Electron A has twice as much momentum as Electron B. Compare their energies.
• EA = EB
• EA = 2 EB
• EA = 4 EB
m
pmvKE
22
1 22
hc
E phand so cpE
double p then quadruple E
double p then double E
21%
47%
33%
22%
40%
38%
MATTER WAVES
de Broglie suggested that Whole of the universe is composed of light (energy) and matter and both are inter-transferable physical quantities (on the basis of Einstein’s mass-energy equivalence relation). Therefore, if light (radiation) possesses the dual character, then matter must show the same, while in motion, as the nature loves symmetry.
E s t i m a t e s o m e d e B r o g l i e w a v e l e n g t h s
• W a v e l e n g t h o f e l e c t r o n w i t h 5 0 e V k i n e t i c e n e r g y2 2
1 02
1 . 7 1 0 m2 2 2e e e
p h hK
m m m K
• W a v e l e n g t h o f N i t r o g e n m o l e c u l e a t r o o m t e m p e r a t u r e
u
1 1
3, M a s s 2 8 m
2
2 . 8 1 0 m3
k TK
h
M k T
• W a v e l e n g t h o f R u b i d i u m ( 8 7 ) a t o m a t 5 0 n K
61 . 2 1 0 m3
h
M k T
Nature of de Broglie waves
Our traditional understanding of a wave….
Adding up waves of different frequencies
A “Wave Packet”
v
Vg
Wave Envelope
Wave Envelope The blue line represents the envelope
function This envelope can travel through space with
a different speed than the individual waves
Wave packet, phase velocity and group velocity ____________________________ The spread of wave packet
in wavelength depends on the required degree of localization in space – the central wavelength is given by
What is the velocity of the wave packet?
p
h
Wave packet, phase velocity and group velocity ________________________________ The velocities of the individual waves which
superpose to produce the wave packet representing the particle are different - the wave packet as a whole has a different velocity from the waves that comprise it
Phase velocity: The rate at which the phase of the wave propagates in space
Group velocity: The rate at which the envelope of the wave packet propagates
Speeds Associated with Wave Packet Speeds Associated with Wave Packet The phase speed of a wave in a wave packet is
given by
This is the rate of advance of a crest on a single wave The group speed is given by
This is the speed of the wave packet itself
phaseωv k
gdωv dk
Wave packet, phase velocity and group velocity ________________________________
Phase velocity
Group velocity
Here is the velocity of light and v is the velocity of the particle
v
cv p
2
vvg
c
Numerical questions related to de Broglie waves
Q An electron and proton have same velocity. Compare the wavelengths, phase velocities and group velocities.
Hint:
Group velocity: vg = v , so group velocity will be same for both.
electron proton= v =
Wavelengths: e = h/mev and p = h/mpv and me < mp so e > p
Phase velocity: vp =c2/v so if v is same then phase velocity will be same for both.
Numerical questions related to de Broglie waves
Q Find the de Broglie wavelength of a 40 keV electron used in certain electron microscope.
Hint: kinetic energy (T) =40 keV = p2/2m, then p = (2mT)1/2
= h/p
Numerical questions related to de Broglie waves
Q Find the de Broglie wavelength of a 1.0 mg grain blown by wind at a speed 20 m / sec.
Hint:
= h / m v = h (1-v2/c2)1/2/ m0 v
Numerical questions related to de Broglie waves
Q. 2
Find the de Broglie wavelength of an electron, whose speed is
(a) 1 x 108 m/sec (b) 2 x 108 m/sec.
Hint:
= h / m v = h (1-v2/c2)1/2/ m0 v
Numerical questions related to de Broglie waves
Q1. A photon and particle have the same de Borglie wavelength. Can anything be said how their linear momentum compare.
Hint: photon = particle
as = h / p therefore, if is same then p will be same for both photon and particle.
Numerical questions related to de Broglie waves
Q Find the KE of an electron whose de-Broglie wave length is same as that of a100 keV x-ray.
HINT: For an x-ray photon E = 100 keV
E=hv=100 keV
hc/ = 100 kev, from this can be determined
Now as given x-ray = electron
Therefore, =h/p implies p = h/
If momentum of the electron is known, then its KE = (½)mv2 = p2/2m
can be determined.
THE UNCERTAINTY PRINCIPLE
In quantum mechanics no measurement can be 100% accurate as with every moving particle a wave packet is associated, hence it is impossible to locate the position and momentum of the particle with 100 % confidence.
Acc to Heisenberg “The product of uncertainty in position of a particle in certain direction and the component of linear momentum of particle in same direction can never be less than ħ/2 or h/4π”
i. e. Δx. Δpx ≥ ħ/2
Wave packet
Δx
THE UNCERTAINTY PRINCIPLE
The narrower is the wave packet more precisely the position of the particle can be determined, however the wavelength in the narrow packet can not be well defined, as there are not enough wavelengths to measure λ accurately. This means that as p=h/λ can not be determined accurately and vice -versa.
Δx is small but Δp is large
Δx is large but Δp is small
Measuring the position and momentum of an electron
Shine light on electron and detect reflected light using a microscope
Minimum uncertainty in position is given by the wavelength of the light
So to determine the position accurately, it is necessary to use light with a short wavelength
Measuring the position and momentum of an electron (cont’d)
By Planck’s law E = hc/λ, a photon with a short wavelength has a large energy
Thus, it would impart a large ‘kick’ to the electron
But to determine its momentum accurately, electron must only be given a small kick
This means using light of long wavelength!
Example of Baseball
A pitcher throws a 0.1-kg baseball at 40 m/s
So momentum is 0.1 x 40 = 4 kg m/s
Suppose the momentum is measured to an accuracy of 1 percent , i.e.,
Δp = 0.01 p = 4 x 10-2 kg m/s
Example of Baseball (cont’d)
The uncertainty in position is then
No wonder one does not observe the effects of the uncertainty principle in everyday life!
Numerical-THE UNCERTAINTY PRINCIPLE (Applications)
Q1. Non-Existence of electron with in the nucleus.
Hint:
let us consider an electron exist
with in nucleus therefore, maximum
Diameter
10-14 m Δx ~x = 10 -14 m
Now by uncertainty principle Δx .Δp ≥ ħ/2, Δp = ħ/2 Δx, Δp = 5.27 x 10 -21 Kgm/sec
As now actual linear momentum p can not be less than Δp (uncertainty in LM) so p ~ Δp Therefore, minimum energy possessed by the electron if it is to stay with in the nucleus is E = p 2/2m 96.4 MeV
Thus, to remain in side the nucleus the minimum energy of the electron should be 96.4 MeV, but it has been found experimentally that no electron can possess energy more than 5 MeV. Thus electron can not exist with in the nucleus.
Numerical-THE UNCERTAINTY PRINCIPLE (Applications)
Q Compare the uncertainties in the velocities of an electron and a proton confined in a 1.00 nm box.
Hint: On the basis of
uncertainty principle
1 x 10 -9 .Δpx ≥ ħ/2,
so Δpx = 0.527 x 10 -25 Kgm/sec for both
Now as we know that Δpx = me Δve , implies that Δve = 0.058 x 106 m/sec
Similarly, Δpx = mp Δvp , implies that Δvp = 0.058 x 102 m/sec
Therefore, Δve > Δvp
Δx= 1.00 nm box
electron proton
Numerical-THE UNCERTAINTY PRINCIPLE (Applications)
Q An electron has a speed of 500 m/sec, correct up to 0.01 %, with what minimum accuracy you can determine the electron.
Hint:
Δvx = (0.01/ 100) x 500 = 0.05 m/sec. 500
Δpx = m Δvx = 9.1 10 -31 x 0.05 = 4.55 10 -32 Kg m Sec-1
Therefore on the basis of uncertainty principle Δx .Δpx ≥ ħ/2
Δx = ħ/ (2 X Δpx) = 2.316 mm.
There is also an energy-time uncertainty relation
Transitions between energy levels of atoms are not perfectly sharp in frequency.
/ 2E t
n = 3
n = 2
n = 1
32E h
32
Inte
nsity
Frequency
32
HEISENBERG UNCERTAINTY PRINCIPLE
There is a corresponding ‘spread’ inthe emitted frequency
810 st
An electron in n = 3 will spontaneouslydecay to a lower level after a lifetimeof order
Wave Function
The wave function is often complex-valued The absolute square |ψ|2 = ψψ is always
real and positive The wave function contains within it all the
information that can be known about the particle
Properties of the wave function When squared, the wave function is a probability density (MaxBorn –
1926). The probability P(x) dx of a particle being between x and x+dx was given in the equation:
The probability of the particle being between x1 and x2 is given by
The wave function must also be normalized so that the probability of the particle being somewhere on the x axis is 1.
Wave Function
The probabilistic interpretation of the wave function The wave function is often complex-valued
The absolute square |ψ|2 = ψψ is always real and positive ψ* is the complete conjugate of ψ
It is proportional to the probability per unit volume of finding a particle at a given point at some instant
The wave function contains within it all the information that can be known about the particle was first suggested by Max Born
Erwin Schrödinger proposed a wave equation that describes the manner in which the wave function changes in space and time
This Schrödinger wave equation represents a key element in quantum mechanics
Wave Function of a Free Particle Because the particle must be somewhere along
the x axis, the sum of all the probabilities over all values of x must be 1
Any wave function satisfying this equation is said to be normalized
Normalization is simply a statement that the particle exists at some point in space
21abP ψ dx
Numerical problems related to probability and normalization
Q Find the value of normalization constant A for the wave function
Hint: Probability of finding the particle equal to 1. square of the wave function should be equal to 1.
Q2 The wave function of a particle is
(a) Find the value of A
(b) Find the probability that the particle be found between x=0 and x=
HINT: (a) for the calculation of A apply normalization condition on the wave function
(b) After substituting the value of A in the wave function find the probability by integrating with in the limits 0 to Pi/2
2/2xAxe
12
0
2/2
dx
x xAxe
2/2/cos2 xforxA
4/
Expectation Values
ψ is not a measurable quantity Measurable quantities of a particle can be derived from ψ The average position is called the expectation value of x and is defined as
The expectation value of any function g(x) for a normalized wave function:
Abbreviated Notation: ‹*| g |›
Operators
Momentum:
Energy:
Expectation Values
Expectation value
the expectation value of momentum involves the representation of momentum
as a quantum mechanical operator.
Where
is the operator for the x component of momentum.
Expectation value
Since the energy of a free particle is given by
and the expectation value for energy becomes
for a particle in one dimension.
Numerical problems related to Expectation values
Q Find the expectation value for the position x of a particle in a box L wide assuming it in the ground state.
Hint: For ground state n=1
L
xn
Lwheredxxx
L sin2
,0
2
L
Schrodinger’s Time dependent equation
Schrodinger equation is the fundamental equation of quantum mechanics in the same sense as the second law of motion of motion is the fundamental equation of classical mechanics
The wave function associated with a moving particle is
The total energy of the sub-atomic particle is E =KE+PE = p2/2m +V
pxEti
Ae
T i m e - i n d e p e n d e n t S c h r ö d i n g e r e q u a t i o n
S u p p o s e p o t e n t i a l i s i n d e p e n d e n t o f t i m e2 2
2i ( )
2V x
t m x
L H S i n v o l v e s o n l y v a r i a t i o n o f Ψ w i t h t
R H S i n v o l v e s o n l y v a r i a t i o n o f Ψw i t h x ( i . e . H a m i l t o n i a n o p e r a t o r d o e s n o t d e p e n d o n t )
L o o k f o r a s e p a r a t e d s o l u t i o n ( , ) ( ) ( )x t x T t
S u b s t i t u t e :
, ( )V x t V x
2 2
2( ) ( ) ( ) ( ) ( ) ( ) ( )
2x T t V x x T t i x T t
m x t
2 2
2 2( ) ( ) ( )
dx T t T t
x d x
2 2
2( )
2
d d TT V x T i
m d x d t
N . B . T o t a l n o t p a r t i a ld e r i v a t i v e s n o w
e t c
Eigenvalues and Eigenfunctions
The values of energy for which the steady state Schrodinder equation can be solved are called eigen values and the corresponding solutions (wave functions) are called as eigen functions.
These words are derived from German words;
EIGEN WERT and EIGEN FUNKTION
eigen means proper or characteristic and wert means value.
To obtain specific values for physical parameters, for example energy, you operate on the wave function with the quantum mechanical operator associated with that parameter
Eigen function and eigen value:
Eigen values Eigen functions
1
2
3
n
1
2
3
n
Numerical problems related to Eigen values and eigen function
Q An eigen function of the operator d2/dx2 is = e3x. Find the corresponding eigen value.
Hint: As we know that Eigen value equation is
here
i = e3x
and Qop = d2/dx2
so when we apply the operator on the wave function the corresponding eigen value comes
qi = 9.
Numerical problems related to Eigen values and eigen function
Q Determine the eigen energy values and eigen functions of a particle trapped in a one dimensional box.
Hint:
2
222
2sin
2
mL
nEand
L
xn
L nn
S U M M A R YS U M M A R Y
T i m e - d e p e n d e n t S c h r ö d i n g e r e q u a t i o n
2 2
2,
2V x t i
m x t
2 *, , , ,P x t d x x t d x x t x t d x
P r o b a b i l i t y i n t e r p r e t a t i o n a n d n o r m a l i z a t i o n
T i m e - i n d e p e n d e n t S c h r ö d i n g e r e q u a t i o n
2 2
2( ) ( ) ( )
2
dV x x E x
m d x
/, ( ) ( ) ( ) i E tx t U x T t U x e
/, ( ) ( ) ( ) i E tx t x T t x e
C o n d i t i o n s o n w a v e f u n c t i o ns i n g l e - v a l u e d , c o n t i n u o u s , n o r m a l i z a b l e ,c o n t i n u o u s f i r s t d e r i v a t i v e
2, , 1d x P x t d x x t
Particle in box:
Particle in a Box
Particle in a Box A particle is confined to a one-
dimensional region of space
The “box” is one- dimensional
The particle is bouncing elastically back and forth between two impenetrable walls separated by L
Potential Energy for a Particle in a Box As long as the particle is inside
the box, the potential energy does not depend on its location
We can choose this energy value to be zero
The energy is infinitely large if the particle is outside the box
This ensures that the wave function is zero outside the box
Wave Function for the Particle in a Box Since the walls are impenetrable, there is zero
probability of finding the particle outside the box
ψ(x) = 0 for x < 0 and x > L
The wave function must also be 0 at the walls ψ(0) = 0 and ψ(L) = 0
Wave Function of a Particle in a Box – Mathematical The wave function can be expressed as a real, sinusoidal
function
Applying the boundary conditions and using the de Broglie wavelength
2( ) sin
πxψ x A
λ
( ) sinnπx
ψ x AL
The Particle in a Box
n is the quantum number (n= 1, 2, 3,....),L is the 'length' of the (one dimensional) molecular box, m is the mass of the particle (electron), and h is Planck's constant.
and
Energy Level Diagram – Particle in a Box The lowest allowed energy
corresponds to the ground state
En = n2E1 are called excited states
E = 0 is not an allowed state
The particle can never be at rest
Graphical Representations for a Particle in a Box
Expectation value for position and linear momentum of the particle in a one dimensional box:
As wave function or eigen function of the particle trapped in a box is known, expectation value of position and momentum can be evaluated from it as:
02
,
22
22222
2
2
.
00
2sin10
2sin2
2
2sin0 21cossin
02
20
20
*0
*
0*2
Ln
Ln
p
ismentummoaverageandforthandbackmovespariclethethus
Ln
mL
nmmEPm
pEthatknowweAs
resultStrange
L
Lxn
Li
L
Lxn
nL
Ln
Liptherefore
axxa
axdxaxSince
dxL
xnCosLL
xnSinL
niL
p
dxL
xnSinLxi
LL
xnSinL
dxxi
LdxpLp
ascomplexlittleispcalculatetobut
dxxLxL
xnSinL
Numerical questions related to Particle in a box:
Q Find the energy of an electron contained in a box of dimension 1 Ao such that its wave function has 5 nodes in the box.
Ans. The energy of en electron is
2
222
2
m
nEn
Numerical questions related to Particle in a box:
Q A particle is described by the wave function
8/1:
2/1,0int
;0
10;3)(
Ans
ervalinfoundbecanparticlethethatyprobabilittheFind
Elsewhere
xxx
Numerical questions related to Particle in a box:
Q Think the nucleus as a box with the size of 10-14 m across. Compute the lowest energy of a neutron confined in the nucleus.
Hint:
considering the nucleus as a cubical box of size 10-14 m.
so that x = y = z = 10-14 m
1
10
2
14
2
2
2
2
2
222
zyx
zyx
z
z
y
y
x
x
nnnand
mhere
nnn
mEn
Numerical questions related to Particle in a box:
Q Show that wave functions for two different states are ortho-normal for a particle in a one dimensional box.
Hint: the wave functions are said to be ortho-normal
if
nmfordx
whennormalizedtosaidarefunctionswavetheand
nmfordx
n
L
m
n
L
m
1
0
0
*
0
*