Question EXE.2.HL.TZ0.17

$\frac{1.24\times {10}^{-6}}{7.80\times {10}^{-11}}$ «= 15.9 keV» ✓

Energy of incident photon =«$15.9\times {10}^3+500=$»$16.4\times {10}^3$ «eV» ✓ 

Wavelength of incident photon =«$\frac{1.24\times {10}^{-6}}{16.4\times {10}^3}=$»$7.56\times {10}^{-11}$ «m» ✓ 

The wavelength of the X-rays changes ✓ 

According to the wave model, the wavelength of the incident and scattered X-rays should be the same ✓

$\left(7.80-7.56\right)\times {10}^{-11}=\frac{1.24\times {10}^{-6}}{0.511\times {10}^6}\left(1-\cos \theta \right)$ ✓

$\theta =89°$ ✓