Q. X-rays of wavelength 0.140 nm are scattered from a block of carbon. What will be the wavelenghts of X-rays scattered at $90^{0}$?
A) 0.140 nm
B) 0.142 nm
C) 0.144 nm
D) 0.146 nm
Explanation
(b) For$\ \phi = 90^{0},\ \cos{\phi = 0}$
So, ${\ \ \lambda}^{'} = \lambda + \frac{h}{m_{e}c}$
= 0.140$\times 10^{- 9} + \frac{6.63 \times 10^{- 34}}{\left( 9.1 \times 10^{- 31} \right)\left( 3 \times 10^{8} \right)}$
= $\left( 0.140 \times 10^{- 9} + 2.4 \times 10^{- 12} \right)\ m = 0.142\ nm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.