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