A photon of wavelength 0.0436 nm strikes a free electron and is scattered at an angle of 34.0 ∘ from its original direction. Part A: Find the change in the wavelength of this photon.
Part B: Find the wavelength of the scattered light.
Part C:Find the change in energy of the photon.
Part D: Find the energy gained by the electron.

a) 4.17*10^-4 nm

b) 0.044 nm

c) 4.35*10^-17 joules

d) 271.875 eV

Explanation:

Given that

angle of scatter = θ = 34°

To find the change in wavelength of the photon, we use the relation

Δλ = λ(f) - λ(i) = [h/me*c] (1 - cos θ)

Δλ = [6.67*10^-34/9.1*10^-31*3*10^8] (1 - cos 34)

Δλ = 2.44*10^-12 * 0.171

Δλ = 4.17*10^-13 m = 4.17*10^-4 nm

B

To find the wavelength of scattered light, we have

Δλ = λ(f) - λ(i) = 4 17*10^-4 nm

λ(f) = λ(i) + 4.17*10^-4 nm, now we substitute the value if λ(i), to get

λ(f) = 4.36*10^-2 nm + 4.17*10^-4 nm

λ(f) = 0.044 nm

C

To find the change in energy, we use our wavelengths in m, instead of nm. The formula is thus

hc/λ(i) - hc/λ(f) = hc[λ(f) - λ(i)]/λ(f) *λ(i)

ΔE = 20.01*10^-26 [4.17*10^-13] / (0.044*10^-9 * 0.0436*10^-9)

ΔE = 8,344*10^-38 ÷ 1.918*10^-21

ΔE = 4.35*10^-17 joules

D

On converting to eV, we have

ΔE = 4.35*10^-17 /1.6*10^-19

ΔE = 271.875 eV (elevtron volt)

-4 on

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