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3-25
E = 300 keV , θ = 30°
(a)
Δλ = λ $ − λ0 =
h
(1 − cos θ ) = (0.002 43 nm )[1 − cos(30°) ] = 3.25 × 10 −13 m
me c
= 3.25 × 10 −4 nm
(b)
(
)(
)
4.14 × 10 −15 eVs 3 × 10 8 m s
hc
hc
E=
⇒ λ0 =
=
= 4.14 × 10 −12 m ; thus,
3
λ0
E0
300 × 10 eV
λ " = λ0 + Δλ = 4.14 × 10 −12 m + 0.325 × 10 −12 m = 4.465 × 10 −12 m , and
(
)(
)
4.14 × 10 −15 eV s 3 × 108 m s
hc
E" =
⇒ E" =
= 2.78 × 10 5 eV .
−12
λ"
4.465 × 10
m
(c)
hc hc
=
+ K e , (conservation of energy)
λ0 λ#
(
)(
)
4.14 × 10 −15 eV s 3 × 10 8 m s
% 1
1(
K e = hc'
−
=
= 22 keV
1
1
& λ0 λ $ *)
−12 −
−12
4.14 ×10
3-28
(a)
λ0 =
(b)
4.465×10
From conservation of energy we have E0 = E " + K e = 120 keV + 40 keV = 160 keV .
hc
The photon energy can be written as E0 =
. This gives
λ0
hc 1 240 nm eV
=
= 7.75 × 10 −3 nm = 0.007 75 nm .
E0 160 × 10 3 eV
Using the Compton scattering relation λ " − λ0 = λ c (1 − cos θ ) where
h
hc 1 240 nm eV
λc =
= 0.002 43 nm and λ " =
=
= 10.3 × 103 nm = 0.010 3 nm .
me c
E" 120 × 10 3 eV
Solving the Compton equation for cos θ , we find
−λc cos θ = λ % − λ 0 − λc
cos θ = 1 −
0.010 3 nm − 0.007 5 nm
λ% − λ0
=1 −
= 1 −1.049 = −0.049
λc
0.002 43 nm
The principle angle is 87.2° or θ = 92.8° .
(c)
Using the conservation of momentum Equations 3.30 and 3.31 one can solve for
the recoil angle of the electron.
p = p" cos θ + pe cos φ
pe sin φ = p# sin θ ; dividing these equations one can solve for the recoil angle of
the electron
tan φ =
€
€
=
and φ = 35.9° .
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p# sin θ
' h*
=) ,
p − p# cos θ ( λ # +
h
λ0
sin θ
' hc *
=) ,
h
− λ # cosθ ( λ # +
120 keV(0.998 8 )
160 keV −120 keV( −0.049)
= 0.723 2
hc
λ0
sin θ
− λ # hc
cosθ
3-30
Maximum energy transfer occurs when the scattering angle is 180 degrees. Assuming the
electron is initially at rest, conservation of momentum gives
2
(me c2 + K )
hf + hf " = pe c =
− m2 c 4 =
( 511 + 50 )2
= 178 keV
while conservation of energy gives hf − hf # = K = 30 keV . Solving the two equations
gives E = hf = 104 keV and hf = 74 keV . (The wavelength of the incoming photon is
hc
λ=
= 0.012 0 nm .
E
3-31
(a)
hc
, λ " = λ0 + Δλ
λ"
6.63 × 10 −34 J ⋅ s 3 × 10 8 m s
hc
λ0 =
=
= 1.243 × 10 −11 m
E0
0.1 MeV
E" =
(
)(
)
(
(
)
)(
6.63 × 10 −34 J ⋅s (1 − cos 60°)
' h *
(
)
Δλ = )
1
−
cos
θ
=
= 1.215 × 10 −12 m
( me c ,+
9.11 × 10 −34 kg 3 × 10 8 m s
λ . = λ0 + Δλ = 1.364 × 10
(
−11
)
m
)(
)
6.63 × 10
J ⋅s 3 × 10 8 m s
hc
E. =
=
= 9.11 × 10 4 eV
λ.
1.364 × 10 −11 m
−34
4-1
F corresponds to the charge passed to deposit one mole of monovalent element at a
cathode. As one mole contains Avogadro’s number of atoms,
96 500 C
e=
= 1.60 × 10 −19 C .
6.02 × 10 23
4-3
Thomson’s device will work for positive and negative particles, so we may apply
q
Vθ
≈ 2 .
m B ld
(a)
4-8
q
Vθ
0.20 radians
7
≈
= (2 000 V)
2 ( 0.10 m )(0.02 m ) = 9.58 × 10 C kg
m B2 ld
−2
4.57 × 10 T
(
)
(b)
As the particle is attracted by the negative plate, it carries a positive charge and is
$ q
'
1.60 × 10 −19 C
=
= 9.58 × 107 C kg)
a proton. &
−27
kg
% mp 1.67 × 10
(
(c)
vx =
(d)
As vx ~ 0.01c there is no need for relativistic mechanics.
(a)
E V 2 000 V
=
=
4.57 × 10 −2 T = 2.19 × 106 m s
B dB 0.02 m
(
)
% sin φ (
From Equation 4.16 we have Δ n∞'
& 2 *)
−4
or Δ n2 = Δ n1
sinφ 1 4
2
sinφ 4
( )
( )
2
2
number of α’s scattered at 40 degrees is given by
. Thus the
4
4
sin 20
(
# sin 10 &
2 )
(
)
Δ n2 = (100 cpm)
=
100
cpm
%$
( = 6.64 cpm .
4
sin 20 '
( sin 402 )
Similarly
Δ n at 60 degrees = 1.45 cpm
Δ n at 80 degrees = 0.533 cpm
Δ n at 100 degrees = 0.264 cpm
(b)
" 1%
2
From 4.16 doubling $ ' mα vα reduces Δ n by a factor of 4. Thus Δ n at
# 2&
" 1%
20 degrees = $ ' (100 cpm) = 25 cpm .
# 4&
(c)
From 4.16 we find
2
Δ nCu ZCu
N
= 2 Cu , ZCu = 29 , ZAu = 79 .
Δ nAu ZAu NAu
NCu = number of Cu nuclei per unit area
= number of Cu nuclei per unit volume * foil thickness
)
# 6.02 × 1023 nuclei & ,
.t = 8.43 × 10 22 t
= + 8.9 g cm 3 %
(
63.54 g
+*
$
' .)
# 6.02 × 10 23 nuclei & ,
22
= + 19.3 g cm 3 %
( .t = 5.90 × 10 t
197.0 g
+*
$
' .-
(
NAu
(
So Δ nCu = Δ n Au ( 29)
4-9
€
)
)
2
8.43 × 10 22
2
(79 )
'
( 5.90 × 10 2 ) = (100)$&% 29
)
79 (
2
$ 8.43 '
&%
) = 19.3 cpm .
5.90 (
The initial energy of the system of α plus copper nucleus is 13.9 MeV and is just the
kinetic energy of the α when the α is far from the nucleus. The final energy of the system
may be evaluated at the point of closest approach when the kinetic energy is zero and the
Ze
potential energy is k( 2e)
where r is approximately equal to the nuclear radius of
r
2
2Ze
€
copper.
Invoking conservation of energy Ei = E f , Kα = ( k )
or
r
2
(
)(
)
)
(2)( 29) 1.60 × 10 −19 8.99 × 109
2Ze 2
r = (k )
=
= 6.00 × 10 −15 m .
Kα
13.9 × 106 eV 1.60 × 10 −19 J eV
€
€
(
)(