The wavelength of the series limit of the Lym | vedclass

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(D) The series limit of the Lyman series corresponds to the transition from $n = \infty$ to $n = 1$ in hydrogen $(Z=1)$.$\frac{1}{\lambda_{\infty}} =

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$37$

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(D) The series limit of the Lyman series corresponds to the transition from $n = \infty$ to $n = 1$ in hydrogen $(Z=1)$.$\frac{1}{\lambda_{\infty}} = R(1)^2 \left( \frac{1}{1^2} - \frac{1}{\infty^2} ight) = R \implies R = \frac{1}{911 \ \mathring{A}}$.For characteristic $K_{\alpha}$ $X$-rays,the minimum wavelength corresponds to the transition from $n = \infty$ to $n = 1$ (Moseley's Law for $K$-series):$\frac{1}{\lambda_{min}} = R(Z - 1)^2 \left( \frac{1}{1^2} - \frac{1}{\infty^2} ight) = R(Z - 1)^2$.Substituting the values:$\frac{1}{0.7 \ \mathring{A}} = \frac{1}{911 \ \mathring{A}} (Z - 1)^2$.$(Z - 1)^2 = \frac{911}{0.7} \approx 1301.4$.Wait,recalculating: The standard value for $R$ is $1.097 	imes 10^7 \ m^{-1}$,and $1/R \approx 911.6 \ \mathring{A}$.$(Z - 1)^2 = \frac{911}{0.7} \approx 1301.4 \implies Z - 1 \approx 36.07$.Thus,$Z \approx 37$.

The wavelength of the series limit of the Lyman series in the hydrogen spectrum is $911 \ \mathring{A}$. Calculate the atomic number of the element that emits characteristic $X$-rays with a minimum wavelength of $0.7 \ \mathring{A}$.

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