Modern physics Chapter 3-4. Solutions to exercises · Modern physics Chapter 3-4. Solutions to...
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Modern physics Chapter 3-4. Solutions to exercises. 3.1.1 The deBroglie wavelength is p h = λ s kgm s kgm h p / 10 89 . 1 / 10 350 10 63 . 6 27 9 34 − − − × ≈ × × = = ⇒ λ 3.2.1 E = hf = eV eV hc 620 10 6 . 1 10 0 . 2 10 0 . 3 10 63 . 6 19 9 8 34 ≈ × × × × × × = − − − λ 3.2.2 With E = eV and p h = λ and mE p m p E 2 2 2 = ⇒ = we get pm m meV h mE h 3 . 12 10 10 10 6 . 1 10 1 . 9 2 10 63 . 6 2 2 3 19 31 34 ≈ × × × × × × × = = = − − − λ 3.2.3 and Å C 7 . 1 = λ kV volt e hc V hc C 3 . 7 10 6 . 1 10 7 . 1 10 00 . 3 10 63 . 6 19 10 8 34 ≈ × × × × × × = = ⇒ = − − − λ λ eV 3.2.4 eV = mv 2 /2; s m s m m eV / 10 21 / 10 1 . 9 10 2 . 1 10 6 . 1 2 2 6 31 3 19 × ≈ × × × × × = = − − v 3.2.5 The deBroglie wavelength Å m meV h m eV m h mv h 39 10 0 . 1 10 6 . 1 10 1 . 9 2 10 63 . 6 2 2 3 19 31 34 ≈ × × × × × × × = = = = − − − − λ 4.1.1 och med Einsteins relation eV 02 , 2 = Φ K hf + Φ = får vi J hc hf K 19 9 8 34 10 60 , 1 02 , 2 10 400 10 00 , 3 10 63 , 6 − − − × × − × × × × = Φ − = Φ − = λ = 4,9725x10 -19 – 2,02x1,60x10 -19 J = 1,7405x10 -19 J J 19 10 7 , 1 − × ≈ 4.1.2 Kinetiska energin 2 2 mv K = ger hastigheten m K v 2 = 4.1.3 K = eV = 1.602x10 -19 J Einstein’s relation hf = Φ + K and with c = fλ we have K hc + Φ = λ which gives = × × − × × × × = − = Φ − − − J K hc 68 . 0 10 602 . 1 10 369 10 00 . 3 10 63 . 6 19 9 8 34 λ 5.12195x10 -19 /1.602x10 -19 – 0.68 eV = 2.5 eV 4.21. The Compton wavelength m m c m h p C 15 8 27 34 10 32 . 1 10 00 . 3 10 673 . 1 10 63 . 6 − − − × = × × × × = = λ
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Transcript of Modern physics Chapter 3-4. Solutions to exercises · Modern physics Chapter 3-4. Solutions to...
Modern physics Chapter 3-4. Solutions to exercises. 3.1.1 The deBroglie wavelength is
ph
=λ skgmskgmhp /1089.1/103501063.6 27
9
34−
−
−
×≈××
==⇒λ
3.2.1 E = hf = eVeVhc 620106.1100.2
100.31063.6199
834
≈×××
×××= −−
−
λ
3.2.2 With E = eV and ph
=λ and mEpm
pE 22
2
=⇒= we get
pmmmeVh
mEh 3.12
1010106.1101.921063.6
22 31931
34
≈××××××
×===
−−
−
λ
3.2.3 and ÅC 7.1=λ kVvolte
hcVhcC
3.7106.1107.1
1000.31063.61910
834
≈×××
×××==⇒= −−
−
λλeV
3.2.4 eV = mv2/2; smsmmeV /1021/
101.9102.1106.122 6
31
319
×≈×
××××== −
−
v
3.2.5 The deBroglie wavelength
ÅmmeVh
meVm
hmvh 39
100.1106.1101.921063.6
22 31931
34
≈××××××
×====
−−−
−
λ
4.1.1 och med Einsteins relation eV02,2=Φ Khf +Φ= får vi
JhchfK 199
834
1060,102,210400
1000,31063,6 −−
−
××−×
×××=Φ−=Φ−=
λ =
4,9725x10-19 – 2,02x1,60x10-19 J = 1,7405x10-19 J J19107,1 −×≈
4.1.2 Kinetiska energin 2
2mvK = ger hastigheten mKv 2
=
4.1.3 K = eV = 1.602x10-19 J Einstein’s relation hf = Φ + K and with c = fλ we have
Khc+Φ=
λ which gives
=××−×
×××=−=Φ −
−
−
JKhc 68.010602.110369
1000.31063.6 199
834
λ
5.12195x10-19/1.602x10-19 – 0.68 eV = 2.5 eV
4.21. The Compton wavelength mmcm
hp
C15
827
34
1032.11000.310673.1
1063.6 −−
−
×=×××
×==λ
4.2.2 E = hf = hc/λ gives pmmEhc 29.8
10602.11015.01000.31063.6
196
834
=××××××
== −
−
λ
4.2.3 The wavelength difference ( ) pmcm
hcm
he
o
e
43.290cos1 ==−=∆λ
4.2.4 K = E – E´ is the kinetic energy.
′−=
′−=
λλλλ11hchchcK
λ = 8.27 pm gives λ´= λ + ∆λ = 9.27 + 2.43 pm = 10.70 pm
keVeVhcK 341070.10
127.81
10602.11000.31063.611 12
19
834
=×
−
××××
=
′−= −
−
λλ
4.3.1 Incoming photon energy hf0= 2 mec2 = 2x511 keV = 1.022 MeV
hf = 2 mec2+ Ke + Kp Kp = 3.0 – 1.022 – 0.25 MeV = 1.73 MeV
