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.

o.022nm

need answer

To find the wavelength of the scattered photon in Compton scattering, we can use the equation that relates the change in wavelength (∆λ) of the photon to the initial wavelength (λ₀) and the scattering angle (θ):

Δλ = λ' - λ₀ = λ₀ * (1 - cosθ)

Where:
- Δλ is the change in wavelength of the photon
- λ' is the wavelength of the scattered photon
- λ₀ is the initial wavelength of the photon
- θ is the scattering angle between the incident and scattered photon

In this case, we are given the energy of the scattered photon (E') as 120 keV and the recoiling electron's energy (Eₑ) as 40 keV. We can use the energy-wavelength equivalence formula for photons:

E = hc/λ

Where:
- E is the energy of the photon
- h is the Planck's constant (6.62607015 × 10⁻³⁴ J·s)
- c is the speed of light (2.998 × 10⁸ m/s)
- λ is the wavelength of the photon

First, let's convert the given energies to joules:

E' = 120 keV = 120 * 1.60218 × 10⁻¹⁹ J
Eₑ = 40 keV = 40 * 1.60218 × 10⁻¹⁹ J

Next, we can find the initial wavelength of the photon (λ₀) using the energy-wavelength equivalence formula:

λ₀ = hc/E'

Then, we can find the wavelength of the scattered photon (λ') using the energy-wavelength equivalence formula for the recoiling electron:

λ' = hc/Eₑ

Finally, we can substitute these values into the Compton scattering equation to find the change in wavelength (∆λ), which will give us the wavelength of the scattered photon (λ'):

∆λ = λ' - λ₀ = λ₀ * (1 - cosθ)

By solving this equation, we can find the wavelength of the scattered photon.

(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