Physics, 14.11.2020 03:00 IntellTanito

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.

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Answer from: andrewsaul04

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

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

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

On converting to eV, we have

ΔE = 4.35*10^-17 /1.6*10^-19

ΔE = 271.875 eV (elevtron volt)

Answer from: Quest

Your answer is 1.5 hope i

Answer from: Quest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

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A photon of wavelength 0.0436 nm strikes a free electron and is scattered at an angle of 34.0 ∘ from...

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