Question

Chapter 38, Problem 036 Your answer is partially correct. Try again. Gamma rays of photon energy 0.563 MeV are directed onto an aluminum target and are scattered in various directions by loosely bound electrons there. (a) What is the wavelength of the incident gamma rays? (b) What is the wavelength of gamma rays scattered at 61.6° to the incident beam? (c) what is the photon energy of the rays scattered in this direction? The electron Compton wavelength is 2.43 × 1012 m (a) NumberTT.02206 (b) NumberT.0348 (c) Number 357004 Units Units Units MeV

Answer 1

A.

We know that

E = h*f = hc/lambda

lambda = hc/E

1 eV = 1.6*10^-19 J

using given values:

lambda = 6.626*10^-34*3*10^8/(0.563*10^6*1.6*10^-19)

lambda = 2.206*10^-12 m = 2.21 pm

B.

d(lambda) = h*(1 - cos phi)/(me*c)

d(lambda) = 6.626*10^-34*(1 - cos 61.6 deg)/(9.31*10^-31*3*10^8)

= 1.244*10^-12 m = 1.244 pm

Now,

d(lambda) = lambda1 - lambda

lambda1 = 1.244 + 2.21 = 3.454 pm