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

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$2.98 \overset{\circ}{A}$

$1.52 \overset{\circ}{A}$

$3.22 \overset{\circ}{A}$

$2.52 \overset{\circ}{A}$

answer is B.

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Detailed Solution

According to Moeley's Law :

$\sqrt{ν} = a (Z - 1)$

$\Rightarrow (Z - 1)^{2} \propto v$

$or (Z - 1)^{2} \propto 1 / λ \Rightarrow$ $\frac{{(Z_{M o} - 1)}^{2}}{{(Z_{C u} - 1)}^{2}} = \frac{λ_{C u}}{λ_{M o}} or λ_{C u} = λ_{M o} \frac{{(Z_{M o} - 1)}^{2}}{{(Z_{C u} - 1)}^{2}}$

$= 0.71 \times {(\frac{41}{28})}^{2} = 1.52 \overset{\circ}{A}$

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