Question

If the $K_{\alpha }$ radiation of $Mo\left(Z = 42\right)$ has a wavelength array of $0.7\overset{o}{A},$ calculate wavelength of the end array corresponding radiation of $Cu,$ i.e., $K_{\alpha }$ for $Cu$ $\left(Z = 29\right)$ assuming $b=1:-$

• A. $2.52\overset{o}{A}$
• B. $1.52\overset{o}{A}$
• C. $2.98\overset{o}{A}$
• D. $3.22\overset{o}{A}$

Answer 1

Answer: (B)

According to Moseley's Law :
$\sqrt{v}=a\left(\right.Z-1\left.\right)$
$\Rightarrow \left(\right.Z-1\left(\left.\right)^{2} \propto v$
or $\left(\right.Z-1\left(\left.\right)^{2} \propto 1/\lambda $
$\Rightarrow $
$\frac{\left(Z_{M o} - 1\right)^{2}}{\left(Z_{C u} - 1\right)^{2}}=\frac{\left(\lambda \right)_{C u}}{\left(\lambda \right)_{M o}}\text{ or }\left(\lambda \right)_{C u}=\left(\lambda \right)_{M o}\frac{\left(Z_{M o} - 1\right)^{2}}{\left(Z_{C u} - 1\right)^{2}}$
$=0.71\times \left(\frac{41}{28}\right)^{2}=1.52\overset{o}{A}$