Matter waves 1 Show that the wavelength of an electron can be expressed as where E is the energy in...

Matter waves

1 Show that the wavelength of an electron can be expressed as

where E is the energy in volts and in nm

1.226

E

34 9

31 19

2 ;( 2 )

6.63 10 1 1.226 10

(2 9.11 10 1.6 10 )

hp mE

me E

E E

Matter waves

2 Show that the wavelength of an electron of energy E can be expressed as

where V is the accelerating potential in volts and in nm

1.50V

V

34 9

31 19

2 ;( 2 )

6.63 10 1 1.50 10

(2 9.11 10 1.6 10 )

hp meV

me V

V V

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) (E in eV)

(b)

1.2260.387 387

10nm pm

1.226

E

3

1.2260.0100 10

15 10nm pm

Matter waves


(c )An electron with energy 1 MeV

(c ) 2 2 2 20

2 2 20

2 2 20

34 8

6 2 3 2 19 2

( ) ( ) sin

1( )

( )

6.63 10 3 10

[(1 10 ) (511 10 ) ] (1.6 10 )

1.45

E pc m c u g relativistic relation

p E m cch hc

p E m c

pm

Matter waves


(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)=(6.6310-34)/(9.1110-31105)

=6.93 10-6m

(iii) =h /p = 6.63 10-34 /[2mE]

=6.6310-34/[2(1.6710-27)(15)1.6 10-

19)]

= 7.410-12 m =7.4pm

Matter waves3 Calculate the de Broglie wavelength of

(iv) A neutron with an energy of 10 keV =h /p = 6.63 10-34 /[2mE]

=6.6310-34/[2(1.6710-27)(10103)1.6 10-19)]= 2.8710-13 m

(v) A neutron moving with a speed of 104 m/s

=h /p = h /mv= 6.63 10-34/ (1.6710-27 ) 104 = 3.8nm

(vi) A ball of 45 g moving with a speed of 22 m/s

=h /p = 6.63 10-34/ 4510-3 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/ 510-3 200 =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

V = 1.5/ 2 = 15 kV

1.50V

V

Davisson and Germer experiment5 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

1.5 1.5( ) 0.0926

175

V Vnm nm

V

d sin =n d=lattice spacing, = wavelength of matter wave; = 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

20( 2 )

hc

K K m c

2 2 2 2 2 2 20 0

2 2 2 2 2 2 20 0 0

2 2 2 20 0 0 0

20

1( ) ( ) ; ( )

( ) ( ) ( )

( )( )

( 2 )

E pc m c p E m cc

h hc hc

p E m c K m c m c

hc

K m c m c K m c m c

hc

K K m c

Wave packet7 Certain ocean waves travel with a phase velocity

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.

2

g

2

( )( ) 1

2( )

1 1

2 2

p

g

p

g gv

k

gd k

d gkd kv gdk dk dk gk

gv

k

Wave packet8 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

=(6.6310-34/(2 10-12)2/(2 9.1110-31)

= 6.03810-20 J=0.377eV

Vg =(2E/m)

=(26.03810-20)/(9.1110-31)

= 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 / =3.31510 -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 ]= 1.28710 -13J

KE= Total energy (E)-Rest energy (m0 c2 )= 4.69810 -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 1.610 -19J)/ 1.28710 -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/[4 210-20] = 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 1010-10 6.63 10-34] = 7.96MHz


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 ≥(400010-10)2/[43108(810-510-10)] = 5.3x 10 –9 s


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 /[45.00 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

(a) E= =2

0n E2 2

28

n h

mL2 34 2

0 2 34 12 2

18

1 0

2 0

[6.6 3 10 ]

8 8 9.11 10 (100 10 )

6.03 10 37.7

4 150.78 ;

9 339.3

hE

mL

J eV

E E eV

E E eV


(b) How much energy must be supplied to excite the electron from the ground state to the first excited state?

(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=1At x=L/2P = (2 / L)[sin2 [( /L)(L/2)] = 2 / LP(dx) = (2/100)10 =0.20

22 2

(1 0) 1 02

0

8[4 1] 37.7 3 113.1

hE n n

mLE eV


(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/3P= (2 / L)[sin2 [( /L)(L/3)] =( 2 / L)(3/4)=1/L

P(dx) =(3/ 2 100)(10) =0.15

At x=L/4P= (2 / L)[sin2 [( /L)(L/4)] =( 2 / L)(1/2)=1/LP(dx) =(1/100)(10) =0.10

At x=LP= (2 / L)[sin2 [( /L)(L)] = (2/L) 0

P(dx)=0


(d) In the first excited state what is the probability of finding the electron between x1 = 0 and x2 = 25 pm?

2

1

2

1

2

2

1

1

21, 2

2( ) {sin ( ) }

1 2{1 cos( ) }

1 2sin( )

2

x

x

x

x

xx

xx

nP x x x dx

L L

nx dx

L L

L nx x

L n L


For the first excited state n=2; Limits X1 =0 and x2 =25pm =L/4

Substituting the value of n and the limits

P =0.25

44

1, 2 00

1 4( ) sin( )

4

1 4 40 sin( ) sin( )

4 4 4 4

1 1 1 1sin sin 0 0 0

4 4 4 4

0.25

LL L

P x x x xL L

L L L L

L L L

L L

L L

Infinite potential well(e) What is the difference in the values of momentum

of the electron in the ground state and 2nd excited state?

2 0 2 0

20 0

0 2

31 18

34

( ) 2 2

2 2

2 1

2 9.11 10 6.01 10 3 1

6.63 10 . /

i P P mE mE

mE n mE

mE n

kg m s

Infinite potential well(e) What is the difference in the values of velocity of

the electron in the ground state and 2nd excited state?

20 02 2

2 0

2 1860

2 31

2 22 2( )

2 2 6.01 101 3 1 7.28 10 /

9.11 10

E n EE Eii v v

m m m m

E nn m s

m m

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?

2 0 22 0 00

10

2 20 0

( )( )2 2 22

1 11 1 1.33 10 133

2 2

h h h hiii

mE mE mEmE n

h hpm

n nmE mE


(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


(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 2

2

2 2

2 10

2sin

/ 2 2 ( )

2 2sin 0

2/ 2 1 ( )

2sin

/ 2 1

2 2sin 2 10

2

nx

L LAt x L and n first excited state

L

L LAt x L and n ground state

nx

L LAt x L and n

L

L L L

Infinite potential well6 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:

Kinetic energy:

Uncertainty:

1 1e ee e

pp

hP

nmhP

2

2

21840

2

e

pe ee p

pp e

p

pmK m

p ppK m

m

21840

2

pe e ee p

p ep p

mv m xx x

v mm x


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

22

0 0 2

34 2

34 9 2

19

22 2

(3 0) 3 02

19

;8

[6.6 3 10 ]

8 9.11 10 (0.251 10 )

9.57 10

8

9.57 10 16 1 90

hE E n E

mL

J

hE n n

mL

eV



(b) What is the difference in the velocity of the electron in the two states ?Solution

20 02 2

2 0

2 1960

2 31

2 22 2( )

2 2 9.57 101 16 1 4.35 10 /

9.11 10

E n EE Eii v v

m m m m

E nn m s

m m



(c )What is the difference in the wavelength of the electron in the two states?

Solution2 0

2 0

200

20

19

10

( )( )2 2

22

11

2

11

162 9.57 10

3.77 10 377

h hiii

mE mE

h h

mEmE n

h

nmE

h

m

m pm



(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 )(mx)

vx ≥ 6.63 10-34/[4 9.1110-31 0.251 10-9]

≥ 230734

Potential barrier8 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 –2k2

L

2 2

31 19 210

34 2

32.4 15

3.24

2 ( )

2 9.11 10 (40 30) 1.6 10 41.618 10

[6.63 10 ]

1

8.5 10

0.1

0.039

m U Ek

For L nm

T e

For L nm

T e

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 T ≈ e –2k

2L

2 2

31 19 29

34 2

1.65

2 ( )

2 9.11 10 (3 0.4) 1.6 10 (2 )8.25 10

[6.63 10 ]

0.1

0.192

m U Ek

For L nm

T e

Matter waves 1 Show that the wavelength of an electron can be expressed as where E is the energy in...

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