Answer
(a) The change in wavelength is the same for both photons.
(b) X-ray
(c) visible: $9.3\times 10^{-4}\%$; X-ray: $16\%$
Work Step by Step
(a) We know that because of scattering, the wavelength of the photon becomes changed. The change in wavelength doesn't depend on the wavelength of the incident photon. Thus, the change in wavelength is the same for both photons.
(b) We know that the incident photon having shorter wavelength experiences greater percent change in wavelength and therefore, the x-ray photon which has the shorter wavelength, experiences the greater percent change in wavelength.
(c) We can find the percentage change in wavelength for the visible light photon as
$\frac{100\Delta \lambda}{\lambda}=\frac{100h}{\lambda m_{\circ}c}(1-cos\theta)$
We plug in the known values to obtain:
$\frac{100\Delta \lambda}{\lambda}=\frac{100h}{\lambda m_e c}(1-cos\theta)$
We plug in the known values to obtain:
$\frac{100\Delta \lambda}{\lambda}=\frac{100(6.34\times 10^{-34}J.s)}{(520\times 10^{-9}m)(9.11\times 10^{-31}Kg)(3\times 10^8m/s)}(1-cos 180^{\circ})=9.3\times 10^{-4}\%$
Now the percentage change in wavelength for the x-ray is given as
$\frac{100\Delta \lambda}{\lambda}=\frac{100h}{\lambda m_e c}(1-cos\theta)$
We plug in the known values to obtain:
$\frac{100\Delta \lambda}{\lambda}=\frac{(100)(6.63\times 10^{-34J.s})}{(0.030\times 10^{-9}m)(9.11\times 10^{-31}Kg)}(1-cos 180^{\circ})=16\%$
