X-rays of wavelength $0.140 \mathrm{~nm}$ are scattered from a block of carbon. What will be the wavelengths…

X-rays of wavelength $0.140 \mathrm{~nm}$ are scattered from a block of carbon. What will be the wavelengths of $X$-rays scattered at $90^{\circ}$ ?
  1. $0.140 \mathrm{~nm}$
  2. $0.142 \mathrm{~nm}$
  3. $0.144 \mathrm{~nm}$
  4. $0.146 \mathrm{~nm}$

Solution

For $\phi=90^{\circ}, \cos \phi=0$
So,
$\begin{aligned}
& \lambda^{\prime}=\lambda+\frac{h}{m_e c} \\
& =0.140 \times 10^{-9}+\frac{6.63 \times 10^{-34}}{\left(9.1 \times 10^{-31}\right)\left(3 \times 10^8\right)} \\
& =\left(0.140 \times 10^{-9}+2.4 \times 10^{-12}\right) \mathrm{m} \\
& =0.142 \mathrm{~nm}
\end{aligned}$

Asked in: AP EAMCET 2008

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