Asked by paul

in the compton scattering event, the scattered photon has an energy of 120 keV and the recoiling e- has an energy of 40 keV. Find the wavelength of the photon.
Answered by muluneh
o.022nm
Answered by remilekun
need answer
Answered by hi
(a)
(b)
(c)
Eγ 2 = 120 keV
Te = 40 keV
Eγ 2
Ee Te
Eγ 1
φ
θ
θ φ
Eγ 1 mec
2 Eγ 2 Te mec
2 + = + +
Eγ 1 = 120 keV 40 keV 160 keV + =
Eγ 1 hν
hc
λ
----- 12 400 eV A˙ ,
λ === --------------------------------
λ 12 400 eV A˙ ,
160 keV = = -------------------------------- 0.0775A˙ ( ) 0.00775nm
λ2 λ1
h
mec ---------( ) 1 – cosθ hc
mec
2 – == = -----------( ) 1 – cosθ 0.0243 1( ) – cosθ
λ2
hc
E2γ
-------- 12 400 eV A˙ ,
120 keV == = -------------------------------- 0.104A˙
( ) 1 – cosθ ( ) 0.103 0.0775 – A˙
0.0243 = = -------------------------------------------- 1.049
cos 0.049 θ = – θ = 92.8°
pγ sinθ pe φ pγ sin ⇒ c sinθ pe = = c sinφ
pec E2
m0
2 – 40 keV 511 keV c4 ( ) + 2 ( ) 511 keV 2 = = –
= 206 keV
sinθ 120 keV 0.999 ⋅
206 keV = = -------------------------------------- 0.582
φ = 35.6
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