Transcript 2 - EngineeringDuniya.com

Matter waves
1 Show that the wavelength of an electron can be expressed as
1.226

E
where E is the energy in volts and  in nm
p  2mE
h
; 
( 2me ) E
6.63 1034
1
1.226 109



31
19
E
E
(2  9.1110 1.6 10 )
Matter waves
2 Show that the wavelength of an electron of energy E can be
expressed as
1.50V

V
where V is the accelerating potential in volts and  in nm
h
p  2meV ;  
( 2me ) V
6.63 10
34
1
1.50 10



31
19
V
(2  9.1110 1.6 10 ) V
9
Matter waves
3 Calculate the de Broglie wavelength of
(i) An electron with energy (a) 10 eV (b) 15 keV (c) 1 MeV
(i)
(a) 

(b)

1.226
E
(E in eV)
1.226
 0.387nm  387 pm
10
1.226
15 10
3
 0.0100nm  10 pm
Matter waves
3 Calculate the de Broglie wavelength of
(c )An electron with energy 1 MeV
(c )
E 2  ( pc)2  (m0c 2 )2
u sin g
relativistic relation
1 2
E  (m0c 2 )2
c
h
hc
 
p
E 2  (m0c 2 )2
p

6.63 1034  3 108
[(1106 )2  (511103 )2 ]  (1.6 1019 ) 2
 1.45 pm
Matter waves
3 Calculate the de Broglie wavelength of
(ii) An electron moving with a speed of 105 m/s
(iii) A proton with an energy of 15 eV
(ii)=(h/p)=(h/mv)=(6.6310-34)/(9.1110-31105)
=6.93 10-6m
(iii)  =h /p = 6.63 10-34 /[2mE]
=6.6310-34/[2(1.6710-27)(15)1.6 10-19)]
= 7.410-12 m =7.4pm
Matter waves
3 Calculate the de Broglie wavelength of
(iv) A neutron with an energy of 10 keV
 =h /p = 6.63 10-34 /[2mE]
=6.6310-34/[2(1.6710-27)(10103)1.6 10-19)]= 2.8710-13 m
(v) A neutron moving with a speed of 104 m/s
 =h /p = h /mv= 6.63 10-34/ (1.6710-27 ) 104 = 3.8nm
(vi) A ball of 45 g moving with a speed of 22 m/s
 =h /p = 6.63 10-34/ 4510-3 22 =6.7 x 10–34 m
(vii) A bullet of 5 g moving with a speed of 200 m/s
 =h /p = 6.63 10-34/ 510-3 200 =6.6 x 10–34 m
Matter waves
4 Compute the accelerating potential required to produce an
electron beam of de Broglie wavelength of 10 pm

1.50V
V
V = 1.5/ 2 = 15 kV
Davisson and Germer experiment
5 A crystal is cut such that the rows of atoms in its surface are separated by a
distance of 0.352nm. A beam of electrons is accelerated through a potential
difference of 175 V and is incident normally on the surface. At what angles
relative to the incident beam would the diffracted beams be observed?
Solution
1.5V
1.5V
 (nm) 

 0.0926nm
V
175
d sin =n 
wave;
d=lattice spacing,  = wavelength of matter
 = angle between the incident & the
scattered beam
sin= ( /d) n ; (0.0926nm/0.352nm)n = 0.263n
 =sin -1 (0.263n) = 15.3, 31.8,52.4 
Matter waves
6 Show that the de Broglie wavelength of a particle of rest mass m0 and kinetic
energy K is given by

hc
K ( K  2 m0 c 2 )
1
E 2  (m0c 2 )2
E  ( pc)  (m0c ) ; p 
c
hc
hc
h

 
p
( K  m0c 2 )2  (m0c 2 ) 2
E 2  (m0c 2 )2
2


2
2 2
hc
( K  m0c 2  m0c 2 )( K  m0c 2  m0c 2 )
hc
K ( K  2m0c 2 )
Wave packet
7 Certain ocean waves travel with a phase velocity
g
2
Where, g is the acceleration due to gravity. Find the group velocity of a
wave packet of these waves in terms of the phase velocity.
vp 
g

2
g
k
g
d (k
)
d ( gk )
d
1
k
vg 



g
dk
dk
dk
2( gk )
1
g 1

 vp
2
k 2
Wave packet
8 An electron has a de Broglie wavelength of 2.00nm. Find its kinetic
energy, phase velocity and the group velocity of the de Broglie
waves.
KE=p2/2m =(h/)2 /2m
=(6.6310-34/(2 10-12)2/(2 9.1110-31)
= 6.03810-20 J=0.377eV
Vg =(2E/m)
=(26.03810-20)/(9.1110-31)
= 3.64 105m/s
vp = c2 /vg =2.5 1011m/s
Wave packet
8 An electron has a de Broglie wavelength of 2.00pm. Find its
kinetic energy, phase velocity and the group velocity of
the de Broglie waves. (use relativistic relation)
p = h /  =3.31510 -22m kg/s; pc = 9.945 10 -14J
E2 = ( pc)2 + (m0 c2 )2 ; =9.89 10 -27+6.684 10 -27
 E = [( pc)2 + (m0 c2 )2 ]= 1.28710 -13J
KE= Total energy (E)-Rest energy (m0 c2 )= 4.69810 -14J
E = m c2 = (m0 c2) /  [1- v2 / c2]
[1- v2 / c2] = (m0 c2)2 / E2 =
 v2 = c2 [1- (m0 c2)2 / E2 ]
 V = c  [1- (m0 c2)2 / E2 ]
= 3 10 8  [1- {(511 10 3 1.610 -19J)/ 1.28710 -13}
=2.3 10 8m /s
Heisenberg uncertainty principle
3 In a gamma decay process the life time of decaying nuclei
is 2 x 10 – 20 s. What is the uncertainty in the energy of
the gamma rays emitted ?
(E)(t) ≥ (h/4) ;  E ≥ h/(4 )(t)
 E ≥ 6.63 10-34/[4  210-20] = 2.6 x 10 –15 J = 16 keV
4 An excited atom radiates a quantum of light of certain
wavelength. Taking 10 ns to be the uncertainty in the time of
radiation calculate the uncertainty in the frequency of the light
emitted. Answer: 7.96 MHz
(E)(t) ≥ (h/4) ;  E ≥ h/(4 )(t) ;
h  ≥ 6.63 10-34/[4  1010-10  6.63 10-34] = 7.96MHz
Heisenberg uncertainty principle
7 If the uncertainty in the location of a particle is equal to its de
Broglie wavelength, what is the uncertainty in its velocity?
Solution:
(px)(x) ≥ (h/4) ;  vx ≥ h/(4 )(m )(x)
vx ≥ h /[4 m] ; vx = (h/ ) / [4 m]= (p/m)/ 4 = v/ 4
8 A spectral line of wavelength 4000Å has a width of 8 10-5 Å. Calculate the
minimum time spent by the electron in the upper energy state between
the excitation and the de-excitation processes.
(E)(t) ≥ (h/4) ; E = hc/ ; E = -(hc/2)  
 t ≥ h 2 /[(4 )(hc)()]

≥ 2 /[(4 )(c)()]
t ≥(400010-10)2/[43108(810-510-10)] = 5.3x 10 –9 s
Heisenberg uncertainty principle
9 A typical atomic nucleus is about 5.00  10–15 m in radius.
Use the uncertainty principle to place a lower limit on the
energy an electron must have if it is to be a part of a
nucleus.
(px)(x) ≥ (h/4) ;  px ≥ h/(4 )(x)
px ≥ h /[45.00 2 10–15];px = 5.275  10–21 kg .m/s
pmin = 5.275  10–21 kg .m/s
 Emin2 = (pc)2 + (m0 c)2 = 1.0023  10–23
 Emin = 3.166  10–12J= 10MeV
Infinite potential well
5 An electron is bound in a one-dimensional potential well of width,
L = 100 pm (a typical atomic diameter), but of infinite wall height.
(a) Find its energy values in the ground state and also in the first two
excited states.
Solution
2 2
(a) E= n h
2
8mL
E0
=
2
n E0
h2
[6.6 3  10 34 ]2


2
8mL
8  9.11 10 34  (100  10 12 ) 2
 6.03  1018 J  37.7eV
E1  4 E0  150.78eV ;
E2  9 E0  339.3eV
Infinite potential well
(b) How much energy must be supplied to excite the electron from
the ground state to the first excited state?
h2
2
2


E(10) 
n

n
0  
2  1
8mL
 E0 [4  1]  37.7  3  113.1eV
(c) Estimate the probability of finding the electron in the ground
state over a stretch of 10 pm each at distances x = L / 2, L /3,
L/ 4 and L from one end of the wall.
Probability density = P =   2 = (2 / L)[sin2 [(n /L)x]
For ground sate n=1
At x=L/2
P = (2 / L)[sin2 [( /L)(L/2)] = 2 / L
P(dx) = (2/100)10 =0.20
Infinite potential well
(c) Estimate the probability of finding the electron in the ground state over a
stretch of 10 pm each at distances x = L / 2, L /3, L/ 4 and L from one
end of the wall.
At x=L/3
P= (2 / L)[sin2 [( /L)(L/3)] =( 2 / L)(3/4)=1/L
P(dx) =(3/ 2  100)(10) =0.15
At x=L/4
P= (2 / L)[sin2 [( /L)(L/4)] =( 2 / L)(1/2)=1/L
P(dx) =(1/100)(10) =0.10
At x=L
P= (2 / L)[sin2 [( /L)(L)] = (2/L) 0
P(dx)=0
Infinite potential well
(d) In the first excited state what is the probability of finding the electron
between x1 = 0 and x2 = 25 pm?
2
P ( x1, x2 ) 
L
1

L
1

L
n
x {sin ( L ) x} dx
1
x2
2
2n
x {1  cos( L ) x}dx
1
x2

L 
2n
x2
 
)x 
 xx1 
sin(
2n 
L
 x1 



x2
Infinite potential well
For the first excited state n=2; Limits X1 =0 and x2 =25pm =L/4
Substituting the value of n and the limits
1  L4 L
P ( x1, x2 )   x0 
L
4


1  L
 L

0



L  4
 4
4  4 

sin( ) x  
L 0 


L
4 L
4 L  

sin(
)

sin(
) 

L 4
L 4 

1 L 1  L 
1   1 
        sin   sin 0       0  0
L  4  L  4 
 4   4 
 0.25
P =0.25
Infinite potential well
(e) What is the difference in the values of momentum
of the electron in the ground state and 2nd excited
state?
i 
( P2  P0 )   2mE2 
2


2mE0 n 



2mE0 
2mE0 

2mE0  n2  1
2  9.11 10
 6.63  10
34
31
 6.01 10
kg .m / s
18
3  1
Infinite potential well
(e) What is the difference in the values of velocity of
the electron in the ground state and 2nd excited
state?
2

 2 E2

2 E0 n
2E0 
2 E2



 ii  (v2  v0 )  

m   m
m 
 m
18
2 E0 n
2  6.0110
6

n

1

3

1

7.28

10
m/ s
2 
 
31
m
9.1110 m
2
Infinite potential well
(e) What is the difference in the values of (iii) de
Broglie wavelength of the electron in the ground
state and 2nd excited state?
 h
h  
h
h 

(iii)(2  0 )  



2
2
mE
2
mE
2mE0 


2
0 
  2mE0 n
h

2mE0
1 
h
  1 
2mE0
 n2 
1 
10
  1  1.33 10  133 pm
 n2 
Infinite potential well
(f) What are the values the uncertainty in the position of the
electron and its momentum?
(px)(x) ≥ (h/4) ;
 px ≥ h/(4 )(x)
px ≥ h /[4  100  10–12] ;
px = 5.28  10–25 kg .m/s
Infinite potential well
(g) What are the values of the probability density at the mid point of the
region, in the ground state and in the first excited state?
Probability density P(x) =
2
sin 2
L
At x  L / 2 and
2
2 L
 sin 2
 0
L
L 2
At x  L / 2 and
2
2
 

sin 2
L
At x  L / 2 and
 

2

n
x
L
n  2 ( first excited
n  1 ( ground
n
x
L
n 1
2
 L 2
sin 2    2 1010
L
L 2 L
state)
state)
Infinite potential well
6 Compare the ground state momenta, the ground state kinetic energy and
the uncertainties in the velocities of an electron and a proton (mass =
1840 x electron mass) confined in a one dimensional “box” of length
1.0 nm.
Momenta:
Pe

Pp
h
h
Kinetic energy:
e
1
p
e  e  1nm
pe 2
mp
Ke
2me


 1840
2
pp
Kp
me
2m p
Uncertainty:
ve

v p
2me xe
2m p x p

mp
me
 1840
xe  x p
pe  p p
Infinite potential well
9
An electron is trapped in an infinitely deep one-dimensional potential well
of width 0.251 nm. Initially electron occupies the n ═ 4 state. Suppose
the electron jumps to the ground state, with the emission of a photon.
(a) what is the energy of the photon?
Solution
2
h
E  E0 n ;
E0 

2
8mL
[6.6 3  10 34 ]2
8  9.11 10 34  (0.251 10 9 ) 2
2
 9.57  10 19 J
h2
2
2


E(3 0) 
n

n
0 
2  3
8mL
19
 9.57  10
16  1  90eV
Infinite potential well
9
An electron is trapped in an infinitely deep one-dimensional potential well
of width 0.251 nm. Initially electron occupies the n ═ 4 state. Suppose
the electron jumps to the ground state, with the emission of a photon.
(b) What is the difference in the velocity of the electron in the two states ?
Solution
2

 2E2

2E0 n
2E0 
2E2



ii  (v2  v0 )  

m   m
m 
 m
19
2E0 n
2  9.57 10
6

n2 1 
16 1  4.35 10 m / s
31
m
9.1110 m
2
Infinite potential well
9
An electron is trapped in an infinitely deep one-dimensional potential well
of width 0.251 nm. Initially electron occupies the n ═ 4 state. Suppose
the electron jumps to the ground state, with the emission of a photon.
(c )What is the difference in the wavelength of the electron in the two states?
Solution


h
h
(iii )(2  0 )  


2 mE2
2 mE0 




 


h
2 mE0 n
2

h
2 mE0

 1

h

1


2 mE0  n2


h




 1


1


 16

2 m  9.57  10 19
 3.77  10 10 m  377 pm
Infinite potential well
9
An electron is trapped in an infinitely deep one-dimensional potential well
of width 0.251 nm. Initially electron occupies the n ═ 4 state. Suppose
the electron jumps to the ground state, with the emission of a photon.
(d) How much is the uncertainty in the velocity of the electron in the potential
well.
Solution
Uncertainty in velocity=vx
(x)(px) ≥ (h/4) ;  vx ≥ h/(4 )(mx)
 vx ≥ 6.63 10-34/[4  9.1110-31 0.251 10-9]
≥ 230734
Potential barrier
8 A 30-eV electron is incident on a square barrier of height 40 eV. What is the
probability that the electron will tunnel through the barrier if its width is (a)
1.0 nm? (b) 0.10 nm?
Solution:
Transmission coefficient T ≈ e –2k2L
k2 
2m(U  E )
2
2  9.111031 (40  30) 1.6 1019  4 2

 1.618  10 10
34 2
[6.63 10 ]
For
L  1nm
T  e  32.4  8.5  1015
For L  0.1nm
T e
 3.24
 0.039
Potential barrier
18 Electrons with energies of 0.400eV are incident on a barrier 3.00eV
high and 0.100nm wide. Find the approximate probability for these
electrons to penetrate the barrier.
Transmission coefficient
k2 
T ≈ e –2k2L
2m(U  E )
2
2  9.111031 (3  0.4) 1.6 10 19  (2 ) 2

[6.63 1034 ]2
For
T e
L  0.1nm
 1.65
 0.192
 8.25 109