Q. The $X$ -ray wavelength of $L_{\alpha }$ line of platinum $\left(\right.Z=78\left.\right)$ is known to be $1.30\overset{o}{A}$ . What is the $X$ -ray wavelength of $L_{\alpha }$ line of molybdenum $\left(\right.Z=42\left.\right)$ ?
NTA AbhyasNTA Abhyas 2020
Solution:
Moseley emperical formula for $L_{\alpha }$ line is given by
$\frac{1}{\lambda }=R\left(\right.Z-7.4\left(\left.\right)^{2}\left(\frac{1}{2^{2}} - \frac{1}{3^{2}}\right)$ .
Given $Z_{1}=78$ , $\lambda _{1}=1.30\overset{o}{A}$ and $Z_{2}=42$ . Let wavelength for $Z_{2}$ is $\lambda _{2}$
then we can write
$\frac{\left(\lambda \right)_{1}}{\left(\lambda \right)_{2}}=\frac{\left(Z_{2} - 7 . 4\right)^{2}}{\left(Z_{1} - 7 . 4\right)^{2}}$ or $\frac{1 . 30}{\left(\lambda \right)_{2}}=\frac{\left(\right. 42 - 7 . 4 \left(\left.\right)^{2}}{\left(\right. 78 - 7 . 4 \left(\left.\right)^{2}}$
$\Rightarrow \lambda _{2}=5.41\overset{o}{A}$
$\frac{1}{\lambda }=R\left(\right.Z-7.4\left(\left.\right)^{2}\left(\frac{1}{2^{2}} - \frac{1}{3^{2}}\right)$ .
Given $Z_{1}=78$ , $\lambda _{1}=1.30\overset{o}{A}$ and $Z_{2}=42$ . Let wavelength for $Z_{2}$ is $\lambda _{2}$
then we can write
$\frac{\left(\lambda \right)_{1}}{\left(\lambda \right)_{2}}=\frac{\left(Z_{2} - 7 . 4\right)^{2}}{\left(Z_{1} - 7 . 4\right)^{2}}$ or $\frac{1 . 30}{\left(\lambda \right)_{2}}=\frac{\left(\right. 42 - 7 . 4 \left(\left.\right)^{2}}{\left(\right. 78 - 7 . 4 \left(\left.\right)^{2}}$
$\Rightarrow \lambda _{2}=5.41\overset{o}{A}$