Compton shift problem

[Brian_D]

Homework Statement

A photon with wavelength 15 pm Compton scatters off an electron, its direction of motion changing by 110 degrees. What fraction of the photon's initial energy is lost to the electron?

Relevant Equations

shift in wavelength=h/mc * (1-cos theta)

h/mc=.00243 nm

Energy of a photon = hc / lambda

(1-cos 110 degrees) is .342, so the Compton shift is .00243 * 10^-9 * .342 = 8.31 * 10^-13 m.

The wavelength of the scattered photon is the Compton shift plus the initial wavelength, that is, (8.31 * 10^-13) + (15 * 10^-12), which equals 1.58 * 10^-11 m. So the energy of the scattered photon is hc / (1.58 * 10^-11), that is, 1.259 * 10^-14 j. And the initial energy of the photon is hc / (15 * 10^-12), that is, 1.362 * 10^-14 j.

Therefore, the fraction of energy lost by the photon should be (1.362 - 1.259) / 1.326, that is, 7.78%. However, the book answer key says 17.9%. If my answer is wrong, what am I doing wrong?

 

[PeroK]

It's better to use algebra and plug the numbers in at the end. You should be able to express the percentage energy loss as a function of $\theta$.

 

[Steve4Physics]

(1-cos 110 degrees) is .342,

$\cos (110^{\circ})$ is negative so you can tell that $1 - \cos (110^{\circ})$ will be greater than 1.

(I haven't checked if that leads to the correct answer though.)

Edit: But @PeroK's suggested approach is the preferred one.

 

[PeroK]

I get 17.8%.

Note that energy is inversely proportional to wavelength.

 

[renormalize]

(1-cos 110 degrees) is .342, so...

This is wrong; check your arithmetic.
EDIT: I also get 17.8%.

 

[Brian_D]

$\cos (110^{\circ})$ is negative so you can tell that $1 - \cos (110^{\circ})$ will be greater than 1.

(I haven't checked if that leads to the correct answer though.)

Edit: But @PeroK's suggested approach is the preferred one.

Yes, thank you Steve4Physics, my mistake was using the wrong value for cos 110 degrees.

 

[Brian_D]

This is wrong; check your arithmetic.
EDIT: I also get 17.8%.

Yes, I see mistake mistake now, and with this correction I also get 18%. Thank you, renormalize.

 

[Brian_D]

It's better to use algebra and plug the numbers in at the end. You should be able to express the percentage energy loss as a function of $\theta$.

Important reminder, thank you PeroK.

 

[PeroK]

Important reminder, thank you PeroK.

The energy of a photon is inversely proportional to the wavelength. Hence
$$\frac{E'-E}{E} = \frac{E'}{E}-1 = \frac{\lambda}{\lambda'} -1$$That is a much simpler calculation.

 

[Brian_D]

Got it, thank you PeroK.