The wavelengths of K_alpha lines of two eleme | vedclass

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(B) According to Moseley's Law for $K_\alpha$ lines: $\frac{1}{\lambda} = R(Z - 1)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = \frac{3R(Z - 1)^2}

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) According to Moseley's Law for $K_\alpha$ lines: $\frac{1}{\lambda} = R(Z - 1)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = \frac{3R(Z - 1)^2}{4}$.Given $R \approx 1.097 	imes 10^7 \, m^{-1}$, so $\frac{1}{R} \approx 912 \, \mathring{A} = 91200 \, pm$.For the first element $(Z_1)$: $(Z_1 - 1)^2 = \frac{4}{3R \lambda_1} = \frac{4 	imes 91200}{3 	imes 250} = \frac{364800}{750} \approx 486.4$. Thus, $Z_1 - 1 \approx 22.05 \Rightarrow Z_1 \approx 23$.For the second element $(Z_2)$: $(Z_2 - 1)^2 = \frac{4}{3R \lambda_2} = \frac{4 	imes 91200}{3 	imes 179} = \frac{364800}{537} \approx 679.3$. Thus, $Z_2 - 1 \approx 26.06 \Rightarrow Z_2 \approx 27$.The elements between $Z_1 = 23$ and $Z_2 = 27$ are $24, 25, 26$. The number of elements is $3$.

The wavelengths of $K_\alpha$ lines of two elements are $250 \, pm$ and $179 \, pm$, respectively. The number of elements between these two elements in the periodic table will be:

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