## College Physics (4th Edition)

(a) $\Delta \lambda = 2.00~pm$ (b) $\lambda_f = 0.152~nm$
(a) We can find the Compton shift in wavelength: $\Delta \lambda = \frac{h}{mc}~(1-cos~\theta)$ $\Delta \lambda = \frac{6.626\times 10^{-34}~J~s}{(9.1\times 10^{-31}~kg)(3.0\times 10^8~m/s)}~(1-cos~80.0^{\circ})$ $\Delta \lambda = (2.427~pm)~(0.82635)$ $\Delta \lambda = 2.00~pm$ (b) We can find the wavelength of the scattered photon: $\Delta \lambda = \lambda_f-\lambda_i$ $\lambda_f = \lambda_i+\Delta \lambda$ $\lambda_f = (0.150~nm)+ (2.00~pm)$ $\lambda_f = (0.150~nm)+ (0.00200~nm)$ $\lambda_f = 0.152~nm$