Question

In a Compton scattering experiment, an x-ray photon scatters through an angle of 16.6

Answer 1

The Compton formula is

lambda' - lambda = C*(1 - cos(theta)),

where lambda' is the wavelength after scattering;
lambda is the wavelength before scattering;
C is the Compton wavelength for an electron (2.43*10^-12 m);
and theta is the angle through which the incident beam is scattered.

We have theta = 17.4 deg. The right side of the equation is then fully determined, and so

lambda' - lambda = 1.11 * 10 ^ (-13) meters.

Now, the energy of a photon is h*c/lambda, so the change in energy of the photon is

h*c*(1/lambda' - 1/lambda) ~ 1.986*10^(-25) kg*m^3/s^2 x (1/lambda' - 1/lambda)

and this must be equal to the energy that the electron _gains_, which is 1/2*me*(2180km/s)^2 ~ 1.265*10^(-18) kg*m^2/s^2.

Therefore, (1/lambda' - 1/lambda) = -1.09*10^7 /meter.

From 1/lambda' - 1/lambda = A and lambda' - lambda = B, we get that lambda' = B + lambda, so that
1/(B+ lambda) - 1/lambda = A, whence we get a quadratic for lambda.

With A = -1.09*10^7 /meter, and B = 1.11 * 10 ^ (-13) meters, we get lambda ~ 1.01*10^(-10) meters (or 0.1 nm, which is smack in the middle of the X-ray wavelength, so we're OK there).

(a) lambda ~ 0.101 nm. Thus, lambda' ~ 0.1011 nm (very little energy is lost).

(b) The photon scatters at 17.4 deg, and carries momentum h/lambda' * sin(17.4 deg) ~ 1.96*10^(-24) kg*m/s in the direction perpendicular to its original motion. This is equal to the electron's perpendicular momentum (negative) after the collision. Thus, me * 2180 km/s * sin (phi) = 1.96*10^(-24) kg*m/s, so that phi ~ 80.7 degrees.