The wavelength of the K_alpha X-ray line for | vedclass

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Question

(A) According to Moseley's law and the Rydberg formula for $X$-ray transitions,the frequency $
u$ is given by $
u = c/\lambda = R(Z-b)^2 (1/n_1^2 -

Vedclass · Accepted Answer

$0.27$

Vedclass · Answer

(A) According to Moseley's law and the Rydberg formula for $X$-ray transitions,the frequency $
u$ is given by $
u = c/\lambda = R(Z-b)^2 (1/n_1^2 - 1/n_2^2)$.For $K_\alpha$ transition,$n_1 = 1$ and $n_2 = 2$. Thus,$1/\lambda_{K_\alpha} \propto (1/1^2 - 1/2^2) = 3/4$.For $K_\beta$ transition,$n_1 = 1$ and $n_2 = 3$. Thus,$1/\lambda_{K_\beta} \propto (1/1^2 - 1/3^2) = 8/9$.Taking the ratio: $\lambda_{K_\beta} / \lambda_{K_\alpha} = (3/4) / (8/9) = (3/4) 	imes (9/8) = 27/32$.Given $\lambda_{K_\alpha} = 0.32 \, \mathring{A}$,we have $\lambda_{K_\beta} = (27/32) 	imes 0.32 \, \mathring{A} = 0.27 \, \mathring{A}$.

The wavelength of the $K_\alpha$ $X$-ray line for a given element is $0.32 \, \mathring{A}$. The wavelength of the $K_\beta$ line emitted by the same element will be ... $\mathring{A}$.

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