Percent change in photon energy

AI Thread Summary
The discussion focuses on calculating the percent change in photon energy after a collision with an electron at rest, specifically for different wavelengths. The equation used is based on the change in wavelength due to scattering, with a specific angle of θ = π/2. Calculations for various scenarios, including microwaves, visible light, X-rays, and gamma rays, yield different percent changes in energy, with notable values like 13.30% for visible light. However, a key point raised is that the original calculations mistakenly focused on wavelength changes instead of energy changes, prompting a request for clarification on the correct method to find the percent change in energy. The conversation emphasizes the importance of correctly interpreting the problem to find the desired energy percentage change.
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1. Homework Statement

A photon of wavelength is incident on an electron at rest. On collision, the photon is scattered at an angle θ with increased wavelength . Suppose θ = pi/2, what is the percent change in photon energy for this collision in (a) a microwave oven with wavelength = 3.0 cm; (b) visible sunlight with wavelength = 5000 Å; (c) X-ray machine with wavelength = 1.0 Å; and (d) ɣ rays with energy 1.0 MeV. [h/mc = 2.426 x 10-12m]

Homework Equations


wavelength’ - wavelength = h/mc (1 – cosϴ)


The Attempt at a Solution



wavelength’ - wavelength = h/mc (1 – cosϴ) = 2.43 x 10-12m (1-cos/2) = 3.76 x 10-4m
a. 3 x 10-2m / 3.76 x 10-4m = 79.79
b. 5000 x 10-10m/ 3.76 x 10-4m = 13.30%
c. 1 x 10-10m / 3.76 x 10-4m = 2.66 x 10-5%
d. 1 MeV = 1 x 106 eV = 1.6 x 10-13 J/ 6.626 x 10-34 Js = 2.41 x 1020 Hz
wavelength = c/f = 2.998 x 108 m/s / 2.41 x 1020 Hz = 1.24 x 10-12 m
1.24 x 10-12 m / 3.76 x 10-4m = 3.30 x 10-7%
( is this correct? please help...)
 
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ikstellub said:
1. Homework Statement

A photon of wavelength is incident on an electron at rest. On collision, the photon is scattered at an angle θ with increased wavelength . Suppose θ = pi/2, what is the percent change in photon energy for this collision in (a) a microwave oven with wavelength = 3.0 cm; (b) visible sunlight with wavelength = 5000 Å; (c) X-ray machine with wavelength = 1.0 Å; and (d) ɣ rays with energy 1.0 MeV. [h/mc = 2.426 x 10-12m]

Homework Equations


wavelength’ - wavelength = h/mc (1 – cosϴ)


The Attempt at a Solution



wavelength’ - wavelength = h/mc (1 – cosϴ) = 2.43 x 10-12m (1-cos/2) = 3.76 x 10-4m

What did you do here? What is cos π/2 equal to? Where did 3.76x10-4 m come from?
a. 3 x 10-2m / 3.76 x 10-4m = 79.79
b. 5000 x 10-10m/ 3.76 x 10-4m = 13.30%
c. 1 x 10-10m / 3.76 x 10-4m = 2.66 x 10-5%
d. 1 MeV = 1 x 106 eV = 1.6 x 10-13 J/ 6.626 x 10-34 Js = 2.41 x 1020 Hz
wavelength = c/f = 2.998 x 108 m/s / 2.41 x 1020 Hz = 1.24 x 10-12 m
1.24 x 10-12 m / 3.76 x 10-4m = 3.30 x 10-7%
( is this correct? please help...)
No, it's not correct. The problem is asking for the percent change in energy. If E and E' are the energy of the photon before and after the collision, what's the expression for the percentage change in energy?
 
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