3-2.
l
r
u
s
!
!
r
r
u
For small values of ! , s " l; therefore, ! =
Recalling that euB =
3-5.
(a)
mu 2
r
" r=
mu !#(2 Ek / e )(e / m )"$
R=
=
qB
(e / m )(B )
mu
eB
s l
"
r r
# !$
l
eBl
=
mu / eB mu
1/ 2
4
1 2 Ek / e
1 " (2 )(4.5 $10 eV / e )#
%
&
=
=
B e/m
0.325T % 1.76 $1011 kg
&(
'
(b)
frequency
M
f =
u
=
2! R
1/ 2
= 2.2 $10!3 m = 2.2mm
(2 Ek / e )(e / m )
2! R
#(2 )(4.5 %104 eV / e )(1.76 %1011 C / kg )$
'
=&
"3
2! (2.2 %10 m )
1/ 2
= 9.1%109 Hz
period T = 1/ f = 1.1%10"10 s
3-6.
(a) 1/ 2mu 2 = Ek , so u =
(2 Ek / e )(e / m )
# u = !%(2 )(2000eV / e )(1.76 $1011 C / kg )"&
(b)
(Problem 3-
"t1 =
1/ 2
x1
0.05m
=
= 1.89 #10!9 s = 1.89ns
7
u 2.65 #10 m / s
(c) mu y = F !t1 = eE!t1
= 2.65 $107 m / s
6 continued)
" u y = (e / m )E#t1 = (1.76 $1011 C / kg )(3.33 $103V / m )(1.89 $10!9 s )= 1.11$106 m / s
3-12.
!mT = 2.898 #10"3 m $ K
(a) !m =
2.898 #10"3 m $ K
= 9.66 #10"4 m = 0.966mm
3K
(b) !m =
2.898 #10"3 m $ K
= 9.66 #10"6 m = 9.66 µ m
300 K
(c) !m =
2.898 #10"3 m $ K
= 9.66 #10"7 m = 966nm
3000 K
3-13. Equation 3-4: R = ! T 4 . Equation 3-6: R =
1
cU .
4
From Example 3-4: U = (8! 5 k 4T 4 )/ (15h3c 2 )
!=
R (1/ 4 )cU 1
=
= c (8" 5 k 4T 4 )/ (15h3c 2T 4 )
4
4
T
T
4
4
=
2" 5 (1.38 $10#23 J / K )
15 (6.63 $10
#34
3
2
J % s ) (3.00 $10 m / s )
3-14. Equation 3-18: u (! ) =
8
= 5.67 $10#8W / m 2 K 4
8" hc! #5
e hc / ! kT # 1
u ( f )df = u (! )d ! # u ( f ) = u ( f )
d!
Because c = f ! ,
df
d!
= c/ f 2
df
5
8" hc ( f / c ) $ c % 8" f 2
hf
u ( f )=
& 2 '= 3
hf / kT
hf / kT
e
(1 ) f *
c e
(1
3-15.
(a) !mT = 2.898 #10"3 m $ K % !m =
2.898 #10"3 m $ K
= 1.07 #10"3 m = 1.07 mm
2.7 K
(b) c = f ! $ f =
c
3.00 #108 m / s
=
= 2.80 #1011 Hz
"3
!m
1.07 #10 m
(c) Equation 3-6:
R=
1
c
cU = (8! 5 k 4T 4 /15h3c3 )
4
4
4
4
(3.00 #10 m / s )(8! )(1.38 #10 J / K ) (2.7 )
=
(4 )(15)(6.63 #10 J $ s ) (3.00 #10 m / s )
8
5
"34
"23
3
3
8
= 3.01#10"6 W / m 2
2
Area of Earth: A = 4! rE2 = 4! (6.38 "106 m )
2
Total power = RA = (3.01#10"6 W / m 2 )(4! )(6.38 #106 m ) = 1.54 #109 W
3-16.
!mT = 2.898 #10"3 m $ K
2.898 "10!3 m # K
= 4140 K
(a) T =
700 "10!9 m
(b) T =
2.898 "10!3 m # K
= 9.66 "10!2 K
3 "10!2 m
2.898 "10!3 m # K
= 9.66 "10!4 K
(c) T =
3m
4
3-17. Equation 3-4: R1 = ! T14
3-18. (a) Equation 3-17: E =
(b) E =
R2 = ! T24 = ! (2T1 ) = 16! T14 = 16 R1
hc / (10hc / kT ) 0.1kT
hc / !
= (hc / kT )/ (10 hc / kT )
= 0.1
= 0.951kT
e
"1 e
"1 e "1
hc / ! kT
hc / (0.1hc / kT ) 10kT
hc / !
=
= 10
= 4.59 #10"4 kT
hc / ! kT
hc / kT )/ (0.1hc / kT )
(
e
"1 e
"1 e "1
Equipartition theorem predicts E = kT . The long wavelength value is very
close to kT, but the short wavelength value is much smaller than the classical
prediction.
3-19. (a) !mT = 2.898 #10"3 m $ K % T1 =
R1 = ! T14
and
2.898 #10"3 m $ K
= 107 K
27.0 #10"6 m
R2 = ! T24 = 2 R1 = 2! T14
! T24 = 2T14 or T2 = 21/ 4 T1 = (21/ 4 )(107 K ) = 128 K
(b) !m =
2.898 #10"3 m $ K
= 23 #10"6 m
128 K
3-20. (a) !mT = 2.898 #10"3 m $ K
!m =
(Equation 3-5)
2.898 #10"3 m $ K
= 1.45 #10"7 m = 145nm
2 #104 K
(b) !m is in the ultraviolet region of the electromagnetic spectrum.
3-21. Equation 3-4: R = ! T 4
Pabs = (1.36 "103W / m 2 )(! RE2 m 2 ) where RE = radius of Earth
Pemit = (RW / m 2 )(4! RE2 )= (1.36 "103W / m 2 )(! RE2 m 2 )
# ! RE2
R = (1.36 %103W / m 2 )&
2
( 4! RE
$ 1.36 %103 W
= "T 4
'=
2
4
m
)
1.36 "103W / m 2
T =
# T = 278.3K = 5.3°C
4 (5.67 "10!8W / m 2 $ K 4 )
4
3-22. (a) !mT = 2.898 #10"3 m $ K % !m =
f m = c / !m =
2.898 #10"3 m $ K
= 8.78 #10"7 m = 878nm
3300 K
3.00 #108 m / s
= 3.42 #1014 Hz
"7
8.78 #10 m
(b) Each photon has average energy E = hf and NE = 40 J / s.
N=
40 J / s
40 J / s
=
= 1.77 "1020 photons / s
!34
14
hf m
(6.63 "10 J # s )(3.42 "10 Hz )
(c) At 5m from the lamp N photons are distributed uniformly over an area
A = 4! r 2 = 100! m 2 . The density of photons on that sphere is (N / A )/ s ! m 2 .
2
The area of the pupil of the eye is ! (2.5 #10"3 m ) , so the number of photons
entering the eye per second is:
2
n = (N / A )(! )(2.5 #10"3 m )
(1.77 #10
=
2
20
2
/ s )(! )(2.5 #10"3 m )
100! m 2
= (1.77 #1020 / s )(! )(2.5 #10"3 m ) = 1.10 #1013 photons / s
3-23. Equation
u (! ) =
3-18:
8" hc! #5
A! #5
Letting
A
=
"
hc
,
B
=
hc
/
kT
,
and
U
!
=
(
)
e hc / ! kT # 1
eB / ! #1
# ! "5 ("1)e B / ! (" B! "2 )
$
du
d # A! "5 $
5! "6 &
%
=
" B/!
%
& = A%
2
B/!
d ! d ! ' eB / ! " 1(
e " 1&
e
"
1
(
)
'
(
"6 B / !
A! "6 # B B / !
$ A! e
#B
$
B/!
=
e
"
5
e
"
1
=
" 5 (1 " e " B / ! )& = 0
(
)
2
2
%
&
%
( (e B / ! " 1) ' !
(
(e B / ! " 1) ' !
The maximum corresponds to the vanishing of the quantity in brackets. Thus,
5! (1 " e " B / ! )= B .
This equation is most efficiently solved by iteration; i.e.,
guess at a value for B/λ in the expression 5! (1 " e " B / ! ), solve for a better value of
B/λ; substitute the new value to get an even better value, and so on. Repeat the
process until the calculated value no longer changes. One succession of values is:
5, 4.966310, 4.965156, 4.965116, 4.965114, 4.965114. Further iterations repeat
the same value (to seven digits), so we have:
6.63 #10"34 J $ s )(3.00 #108 m / s )
(
B
hc
hc
= 4.965114 =
% !mT =
=
!m
! m kT
(4.965114 )k (4.965114 )(1.38 #10"23 J / K )
!mT = 2.898 #10"3 m $ K
3-26. (a) "t =
(Equation 3-5)
hc 1240eV $ nm
=
= 653nm,
!
1.9ev
ft =
!
1.9eV
=
= 4.59 %104 Hz
h 4.136 %10#15 eV $ s
1 $ hc
% 1 $ 1240eV # nm
%
& 1.9eV ( = 2.23V
(b) V0 = ' & ! ( = '
e) "
* e ) 300nm
*
1 $ hc
% 1 $ 1240eV # nm
%
& 1.9eV ( = 1.20V
(c) V0 = ' & ! ( = '
e) "
* e ) 400nm
*
3-27. (a) Choose λ =550nm for visible light. nhf = E !
dn
dE
hf =
=P
dt
dt
(0.05 #100W )(550 #10"9 m )
dn P P!
=
=
=
= 1.38 #1019 / s
"34
dt hf
hc (6.63 #10 J $ s )(3.00 #10 m / s )
(b) flux =
number radiated / unit time 1.38 "1019 / s
=
= 2.75 "1017 / m 2 # s
2
area of the sphere
4! (2m )
3-28. (a) hf = ! # ft =
(b) f = c / ! =
!
4.22eV
=
= 1.02 $1015 Hz
"15
h 4.14 $10 eV % s
3.00 #108 m / s
= 5.36 #1014 Hz
"9
560 #10 m
No.
Available energy/photon hf = (4.14 "10!15 eV # s )(5.36 "1014 Hz )= 2.22eV .
This is less than ! .
3-32. (a) ! =
hc 1240eV nm
=
= 1.90eV
"
653nm
(b) Ek =
hc
1240eV nm
#! =
# 1.90eV = 2.23eV
"
300nm