5-3. ( ) ( )

( )( )

2 22

22 2 2 5

12401940

2 2 2 5 11 10 0 04. .k o o

hc eV nmpE eV V V

m mc e eV nm! != = = = =

"

��

5-4. 2

(from Equation 5-2)2 2k k

h h hc

p mE mc E! = = =

(a) For an electron: ( )( )( )

1 26 3

12400 0183

2 0 511 10 4 5 10/

.

. .

eV nmnm

eV eV

! = =" #$ $% &

�

(b) For a proton: ( )( )( )

4

1 26 3

12404 27 10

2 983 3 10 4 5 10/

.

. .

eV nmnm

eV eV

! "= = #$ %# #& '

�

(c) For an alpha particle:

( )( )( )4

1 29 3

12402 14 10

2 3 728 10 4 5 10/

.

. .

eV nmnm

eV eV

! "= = #$ %# #& '

�

5-5. ( )

1 222 2 1 5 (from Equation 5-2)

/

/ / / .k

h p h mE hc mc kT! " #= = = $ %

Mass of N2 molecule = ( )2 4 2 10 2

2 14 0031 931 5 2 609 10 2 609 10. . / . / . /u MeV uc MeV c eV c! = ! = !

( )( )( )( )( )

1 210 5

12400 0276

2 2 609 10 1 5 8 617 10 300/

.

. . . /

eV nmnm

eV eV K K

!"

= =# $% %& '

�

5-11. 0 25

2

.

p k

h hnm

p m E! = = =

Squaring and rearranging,

( )

( )( )

( )( )

2 22

22 2 2 6

12400 013

2 2 2 938 10 0 25

.

.k

p p

hc eV nmhE eV

m m c eV nm! != = = =

"

�

( )( ) ( )1 0 25 0 304sin sin / . / .n D n D nm nm! " " != # = =

0 822 55sin .! != " = °

5-12. (a) 2

2

sinsin sin

k

n nhcn D D

mc E

!! "

" "= # = =

( )( )

( ) ( )( )1 2

5

1 12400 210

55 6 2 5 11 10 50/

.

sin . .

eV nmnm

eV eV

= =! "° #$ %

�

(b) ( )( )

( ) ( )( )1 2

5

1 12400 584

0 210 2 5 11 10 100/

sin .

. .

eV nmn

Dnm eV eV

!" = = =

# $%& '

�

( )10 584 35 7sin . .! "= = °

5-17. (a) 1 2

y y y= + ( ) ( )0 002 8 0 400 0 002 7 6 380. cos . / / . cos . / /m x m t s m x m t s= ! + !

( ) ( ) ( )1 1

2 0 002 8 0 7 6 400 3802 2

. cos . / . / / /m x m x m t s t s! "

= # # #$ %& '

( ) ( )1 18 0 7 6 400 380

2 2cos . / . / / /x m x m t s t s

! "# + $ +% &

' (

( ) ( )0 004 0 2 10 7 8 390. cos . / / cos . / /m x m t s x m t s= ! " !

(b) 39050

7 8

//

. /

sv m s

k m

!= = =

(c) 2050

0 4

//

. /s

sv m s

k m

!"= = ="

(d) Successive zeros of the envelope requires that 0 2 , . /x m !" = thus

1

1 2

25 with 0 4 and 5

0 2. .

.

x m k k k m x mk

! !! !"

# = = # = " = # = =#

5-22. (a) ( )( )

1 22 6

12400 549

2 2 2 0 511 10 5/

.

.k k

h h hc eV nmnm

p mE mc E eV eV

! = = = = =" #$% &

�

2 For first minimum (see Figure 5-17).sin /d ! "=

0 5493 15 slit separation

2 2 5

..

sin sin

nmd nm

!

"= = =

°

(b) 0 55 0 5 where = distance to detector plane 5 74

2 5

.sin . / .

sin

cmcm L L L cm° = = =

°

5-23. (a) The particle is found with equal probability in any interval in a force-free

region.

Therefore, the probability of finding the particle in any interval ∆x is

proportional to

∆x. Thus, the probability of finding the sphere exactly in the middle, i.e.,

with

∆x = 0 is zero.

(b) The probability of finding the sphere somewhere within 24.9cm to 25.1cm is

proportional to ∆x =0.2cm. Because there is a force free length L = 48cm

available

to the sphere and the probability of finding it somewhere in L is unity, then

the

probability that it will be found in ∆x = 0.2cm between 24.9cm and 25.1cm

(or any

interval of equal size) is: ( )( )1 48 0 2 0 00417/ . . .P x cm! = =

5-24. Because the particle must be in the box ( )2 2

0 0

1 1* sin /

L L

dx A x L dx! ! "= = =# #

Let ( )0 0 and / ; ; /u x L x u x L u dx L du! ! != = " = = " = = , so we have

( ) ( )2 2 2 2

0 0

1/ sin / sinA L udu A L udu

! !

! != =" "

( ) ( ) ( ) ( ) ( )2 2 2 2 2

00

22 2 1

2 4

sin/ sin / / / /

u uL A udu L A L A LA

!!

! ! ! !" #

= $ = = =% &' (

)

( )1 22

2 2/

/ /A L A L! = " =

5-25. (a) At ( )22

0 20 0 0: ,x Pdx dx Ae dx A dx!= = = =

(b) At 2 2

2 24 1 4 2

0 61/ /

: .x Pdx Ae dx Ae dx A dx! !! " "

= = = =

(c) At 2 2

2 24 4 1 2

2 0 14/

: .x Pdx Ae dx Ae dx A dx! !! " "

= = = =

(d) The electron will most likely be found at x = 0, where Pdx is largest.

5-27. ( )

34

9

7 19

1 055 106 6 10

10 1 609 10

./ .

. /

J sE t E t eV

s J eV

!!

! !

"# # $ % # $ # = $ "

"

�h h

5-35. The size of the object needs to be of the order of the wavelength of the 10MeV

neutron.

and are found from:/ / .h p h mu u! " "= =

( )21 or 1 10 939/

k nE m c MeV MeV! != " " =

( )1 2

2 21 10 939 1 0106 1 1 or 0 14

/

/ . / / .u c u c! = + = = " =

Then, ( ) ( )( )( )2 6

12409 33

1 0106 939 10 0 14

./ . .

h hc eV nmfm

mu mc u c eV!

" "= = = =

# $ # $%& ' & '

�

Nuclei are of this order of size and could be used to show the wave character of 10MeV neutrons.

5-39. 2 2

E t tE

! ! " # ! =!

h h

16

24

6

6 58 101 32 10

2 250 10

..

eV st s

eV

!!"

# = = "" "

�

5-40. (a) For a proton or neutron:

and 2

x p p m v! ! " ! = !h assuming the particle speed to be non-relativistic.

( )( )

34

7

27 15

1 055 103 16 10 0 1

2 2 1 67 10 10

.. / .

.

J sv m s c

m x kg m

!

! !

"# = = = " $

# "

h � (non-

relativistic)

(b) ( )( )

227 7

2 131 67 10 3 16 101

8 34 10 5 212 2

. . /. .

k

kg m sE mv J MeV

!

!" "

# = = " =

(c) Given the proton or neutron velocity in (a), we expect the electron to be

relativistic,

in which case, ( )21 and

kE mc != "

2 2

p mv vx m x

! !" = # $ #" "

h h

For the relativistic electron we assume v c!

( )( )( )

34

31 8 15

1 055 10193

2 2 9 11 10 3 00 10 10

.

. . /

J s

mc x kg m s m!

"

" "

#$ = =

% # #

h �

( ) ( )( ) ( )2

2 31 8 111 9 11 10 3 00 10 192 1 58 10 98. . / .

kE mc kg m s J MeV! " "= " = # # = # =