X-rays of wavelength $0.140 \mathrm{~nm}$ are scattered from a block of carbon. What will be the wavelengths…
- $0.140 \mathrm{~nm}$
- $0.142 \mathrm{~nm}$
- $0.144 \mathrm{~nm}$
- $0.146 \mathrm{~nm}$
Solution
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