Question

X-rays of energy 50 KeV are scattered from a carbon target. The scattered rays are at ${90}^o$ from the incident beam. The percentage of change in wavelength is
(given $m_e=9\times {10}^{-31}Kg,C=3\times {10}^8$m/s)

• A. 10%
• B. 20%
• C. 5%
• D. 1%

Answer 1

Answer: (A)

10%

$\theta ={90}^{\circ }$
so, $cos\theta =0$
$\Delta \lambda =\frac{h}{m_{e}C}(1-cos\theta )=\frac{h}{m_{e}C}(1-0)=\frac{h}{m_{e}C}$

percentage of change in wavelength

$\frac{\Delta \lambda }{{\lambda }_i}\times 100$ % $(\Delta \lambda =\frac{h}{m_{e}C})$

$=\frac{h/m_{e}c}{hc/energy}\times 100(energy=\frac{hc}{\lambda })$

$=\frac{energy}{m_e{C}^2}\times 100$

$=\frac{50\times {10}^3\times 1.6\times {10}^{-19}\times 100}{9\times {10}^{-31}\times 3\times {10}^8\times 3\times {10}^8}$

$=1\times 10$
$=10$%

So, the answer is option (A).