Problem 4 a) An incident 71 pm X-ray is incident on a calcite target. i. Find the wavelength of the X-ray scattered at a 30° angle, ii. What is the largest shift that can be expected in this experiment? b) X-rays of wavelength 10.0 pm are scattered from a target. i. Find the wavelength of x-rays scattered through 45° ii. Find the maximum wavelength present in the scattered x-rays. iii. Find the maximum kinetic energy of the recoil electrons, c) At what scattering angle will incident 100 keV x-rays leave a target with an energy of 90 keV.
Answers
Answer:
Hi paingomaima. I hope this answer would help you
Compton shift is given as Δλ=
mc
h
(1−cosϕ)=
9.1×10
−31
Kg×3×10
8
m/s
6.62×10
−34
Js
(1−cos85
0
)
where m is the mass of electron and h is Planck's constant with c as speed of light in air.
On calculation we get the shift as Δλ=2.2×10
−12
meter=2.2picometer
Answer:
Explanation:
From the above question,
They have given :
a) i. The wavelength of the X-ray scattered at a 30° attitude can be observed the use of the Bragg equation:
2d sin θ = nλ
For calcite, d = 2.71 Å. Converting the incident wavelength to Å,
we have:
λ = 0.071 nm
= 0.71 Å
Substituting these values into the Bragg equation, we get:
2(2.71 Å) sin 30° = 1λ
λ = 1.24 Å
Therefore, the wavelength of the X-ray scattered at a 30° attitude is 1.24 Å.
ii. The greatest shift that can be anticipated in this scan is given through the Compton shift:
Δλ = λ(1 - cosθ)
where λ is the incident wavelength and θ is the scattering angle.
Substituting the values we have:
Δλ = 0.71 Å (1 - cos 30°)
Δλ = 0.118 Å
Therefore, the greatest shift that can be anticipated in this test is 0.118 Å.
b) i. The wavelength of x-rays scattered via 45° can additionally be determined the use of the Bragg equation, assuming that the goal cloth has atomic planes separated with the aid of d = λ/2:
2d sin θ = nλ
Substituting the given values:
2(λ/2) sin 45° = 1λ'
λ' = λ/√2
λ' = 7.07 pm
Therefore, the wavelength of x-rays scattered thru 45° is 7.07 pm.
ii. The most wavelength current in the scattered x-rays is given via the Compton shift:
Δλ = λ(1 - cosθ)
Substituting the given values:
Δλ = 10.0 pm (1 - cos 180°)
Δλ = 20.0 pm
Therefore, the most wavelength existing in the scattered x-rays is 20.0 pm.
iii. The most kinetic strength of the shy away electrons is given via the Compton formula:
E = h(c/λ - c/λ')
where h is Planck's constant, c is the pace of light, λ is the incident wavelength, and λ' is the scattered wavelength.
Substituting the given values:
E = h(c/10.0 pm - c/20.0 pm)
E = 1.24 keV
Therefore, the most kinetic power of the pull away electrons is 1.24 keV.
c) The electricity of the scattered X-rays is given by:
E' = E/(1 + E/mc^2*(1-cosθ))
where E is the incident energy, m is the relaxation mass of the electron, and c is the velocity of light.
Substituting the given values:
E = one hundred keV
E' = ninety keV
θ = ?
Solving for θ, we get:
cos θ = 1 - (mc^2(E - E'))/(E'E + mc^2*E')
cos θ = 0.954
θ = 18.26°
Therefore, the scattering attitude is about 18.26°.
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