Photon Scattering: Find Wavelength of Incident Photon

[Fizzicist]

Homework Statement

A photon scatters in the backward direction ($$\theta$$= 180) from a free proton that is initially at rest.

What must the wavelength of the incident photon be if it is to undergo a 10.0% change in wavelength as a result of the scattering?

Homework Equations

$$\lambda$$'-$$\lambda$$ = (h/mc)(1-cos($$\theta$$))

where the left side is the difference between scattered and incidence wavelengths.

The Attempt at a Solution

This seemed like a pretty straightforward problem. Since the photon undergoes a 10% change in wavelength, 1.1$$\lambda$$ = $$\lambda$$'. Therefore .1$$\lambda$$ = (h/mc)(1-cos($$\theta$$)). Multiply by 10 and evaluate the cosine, and you get $$\lambda$$ = 20h/mc. However, when I substitute values into this and evaluate it I get the wrong answer. I have absolutely no clue what I am doing wrong here. This shouldn't be a difficult problem, but for some reason I am not getting the correct answer. Help would be appreciated. Thanks.

 

[astrorob]

Hey,

Could you post a bit more of the working, as in the actual values you're putting into evaluate it? & the answer if you have it.

 

[Fizzicist]

Hey,

Could you post a bit more of the working, as in the actual values you're putting into evaluate it? & the answer if you have it.

Sure. Here's my work:

$$\lambda$$'-$$\lambda$$ = (h/mc)(1-cos$$\theta$$)

1.1$$\lambda$$-$$\lambda$$ = (1 - cos180)(h/mc)

.1$$\lambda$$ = 2h/mc

$$\lambda$$ = 20h/mc = 20 * (6.626 * 10^-34)/(9.109 * 10^-31)(3.00 * 10^8) = .04852 nm

 

[Fizzicist]

Does anyone know what I'm doing wrong here?

 

[alphysicist]

Hi Fizzicist,

You used the mass of an electron, but in this problem the scattering is from a proton.

 

[Fizzicist]

d'oh! haha...thanks...


solved.