Question

$X$-rays of wavelength $0.140nm$ are scattered from a block of carbon. What will be the wave lengths of $X$-rays scattered at $90$ ?

• A. $0.140nm$
• B. $0.142nm$
• C. $0.144nm$
• D. $0.146nm$

Answer 1

Answer: (B)

$0.142nm$

For $\varphi =90,\cos \varphi =0$
So, ${\lambda }^{\prime }=\lambda +\frac{h}{m_{e}c}$
$=0.140\times {10}^{-9}+\frac{6.63\times {10}^{-34}}{9.1\times {10}^{-31}3\times {10}^8}$
$=(0.140\times {10}^{-9}+2.4\times {10}^{-12})m$
$=0.142nm$