## Related questions with answers

(a) Calculate the ratio of $\mathrm{K}_{\alpha}$ wavelengths for uranium and carbon. (b) Calculate the ratio of $\mathrm{L}_{\alpha}$ wavelengths for platinum and calcium.

Solution

VerifiedExpression for $K_{\alpha}$ is

$\begin{align*} \lambda_{K_{\alpha}} &= \frac{4}{3 R} \frac{1} { \left( Z - 1 \right)^2} \tag{1}. \end{align*}$

We are asked to calculate ratio of $K_{\alpha}$ wavelengths for uranium and carbon, which have atomic number of 92 and 6. So, using eqaution (1), we get

$\begin{align*} \frac{ \lambda_{K_{\alpha}, U}}{ \lambda_{K_{\alpha}, C}} &= \frac{\frac{1} { \left( 92 - 1 \right)^2}}{\frac{1} { \left( 6 - 1 \right)^2}} \\ \frac{ \lambda_{K_{\alpha}, U}}{ \lambda_{K_{\alpha}, C}} &= \frac{5^2}{91^2} \\ \frac{ \lambda_{K_{\alpha}, U}}{ \lambda_{K_{\alpha}, C}} &= \boxed{3.02 \cdot 10^{-3} } \end{align*}$

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