PHYSICS 4E
PROF. HIRSCH
QUIZ 3
SPRING QUARTER 2011
MAY 9th
Formulas and constants:
hc = 12,400 eV A ; k B = 1/11,600 eV/K ; ke 2 = 14.4eVA ; me c 2 = 0.511"10 6 eV ; m p /me = 1836
m2 c 4 + p 2 c2
;
c = 3 ! 108 m / s
r
r
r r
Lorentz force: F = qE + qv " B
_
_
8#
hc
1
Planck's law : u( ") = n( " ) E ( ") ; n( ") = 4
;
E ( ") =
hc / "kB T
"
" e
$1
#E / kB T
Energy in a mode/oscillator : E f = nhf ; probability P(E) " e
!
!
&
Relativistic energy - momentum relation E =
Photons: E = hf ; p = E /c ; f = c / "
!
Stefan's law : R = "T 4 ; " = 5.67 #10$8 W /m 2K 4 ; R = cU /4 , U =
!
!
0
8#
u( " ) = n( ") E ( " ) ; n( " ) = 4
"
_
Planck :
hc
1
; E (") =
; Wien : "m T = hc /4.96k B
hc / "kB T
" e
$1
_
1
Photoelectric effect : eV0 = ( mv 2 ) max = hf " # ,
2
!
!
Compton : "2 - "1 =
# $ work function
h
h
kq Q
1
(1# cos$ ) ; "c %
= 0.0243A ; Rutherford : b = & 2 cot($ /2) ; 'N ( 4
mec
mec
m& v
sin ($ /2)
kq1q2
kq
(force) ; V =
(potential) ; U = q0V (potential energy)
2
r
r
1
1
1
1
Hydrogen spectrum :
= R( 2 # 2 )
;
R = 1.097 $10 7 m#1 =
"
m
n
911.3A
Electrostatics : F =
!
!
!
!
!
!
Bohr atom : rn = r0 n 2 ; r0 =
a0
Z2
h2
ke 2
; E n = "E 0 2 ; a0 =
=
0.529A
;
E
=
= 13.6eV ; L = mvr = nh
0
Z
n
mke 2
2a0
1
ke 2 Z
ke 2 Z
v2
E k = mv 2 ; E p = "
; E = Ek + E p ; F = 2 = m
; hf = hc/# = E n " E m
2
r
r
r
mM
Reduced mass : µ =
; X - ray spectra : f 1/ 2 = An (Z " b) ; K : b = 1, L : b = 7.4
m+M
h
E
2$
p2
de Broglie : " = ; f =
; # = 2$f ; k =
; E = h# ; p = hk ; E =
; hc = 1973 eV A
p
h
"
2m
!
Wave packets : y(x,t) = $ a j cos(k j x " # j t), or y(x,t) =
!
group and phase velocity : v g =
!
!
!
!
!
!
!
!
!
' u(%)d%
% dk a(k) e
i(kx -# (k )t )
; &k&x ~ 1 ; &#&t ~ 1
j
d"
dk
;
vp =
"
k
;
Heisenberg
: #x#p ~ h
#t#E ~ h
;
"(x,t) =| "(x,t) | e i# (x,t ) ;
P(x,t) dx =| "(x,t) |2 dx = probability
E
-i t
h2 " 2#
"#
h
Schrodinger equation : +
V(x)#(x,t)
=
ih
;
#(x,t)
=
$
(x)e
2m "x 2
"t
%
h 2 # 2$
Time " independent Schrodinger equation : +
V(x)
$
(x)
=
E
$
(x)
;
& dx $ *$ = 1
2m #x 2
-%
Wave function
" square well : # n (x) =
2
n$x
$ 2h2n 2
sin(
) ; En =
;
L
L
2mL2
x op = x , pop =
h %
; < A >=
i %x
Eigenvalues and eigenfunctions : Aop " = a " (a is a constant) ; uncertainty :
Harmonic oscillator : "n (x) = Cn H n (x)e
(k1 " k2 ) 2
Step potential : R =
,
(k1 + k2 ) 2
#
m$ 2
x
2h
"
*
& dx# A
#
op
-"
#A = < A 2 > $ < A > 2
1
p2 1
1
; E n = (n + )h$ ; E =
+ m$ 2 x 2 = m$ 2 A 2 ; %n = ±1
2
2m 2
2
T = 1" R
;
k=
2m
(E " V )
h2
PHYSICS 4E
PROF. HIRSCH
QUIZ 3
SPRING QUARTER 2011
MAY 9th
b
%
-2 # (x )dx
" (x) ~ e -#x
Tunneling :
;
T ~ e -2#$x ;
T~e
a
;
# (x) =
Justify all your answers to all problems
2m[V (x) - E]
h2
Problem 1 (10 pts)
An electron in an infinite one-dimensional square well has energy 5 eV.
(a) What is the smallest possible width of this well? Give your answer in A. What are two
other possible value for the width of this well?
!
Assuming the width of the well is the smallest found in (a),
(b) What is the uncertainty in the momentum of this electron, Δp? Give your answer in
units eV/c. Do not use Heisenberg's uncertainty principle to find this answer, use the
definition of uncertainty given in the formula sheet.
(c) To check whether your answer in (b) makes sense, do the following: if L is the width
of the well, you can assume that Δx~L, so by the uncertainty principle ΔpL should equal
a dimensionless number (of order unity) times h . Find that dimensionless number.
(d) How much more likely is it to find this electron in a small region around x=L/2 than
in a small region of the same width around x=L/6?
Use: h 2 /2me = 3.81 eVA 2
!
!
!
Problem 2 (10 pts)
The wavefunction for an electron in a harmonic oscillator potential is
2
2
" (x) = Ce#x / a with a=1A.
(a) Find C, in units A-1/2.
(b) Find the uncertainty in the position, Δx, give its value in A.
(c) Find the energy of this electron, in eV.
$
$
2
2
&
1 &
Use % dx e" #x =
, % dx x 2e" #x =
# "$
2# #
"$
Problem 3 (10 pts)
!
2A
!
10,000 electrons incident
100 electrons
1 electron
V0
1eV
x=0
10,000 electrons are incident from the left. Only 100 electrons make it past the potential
step of height 1 eV at x=0, and only 1 electron makes it past the barrier of height V0 and
width 2A.
(a) Find the kinetic energy of the incident electrons, in eV.
(b) Find the value of V0 in eV.
Justify all your answers to all problems