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If $λ_{C u}$ is the wavelength of $K_{α}$ X-ray line of copper (atomic number 29) and $λ_{M o}$ is the wavelength of the $K_{α} X$ -ray line of molybdenum (atomic number 42) then the ratio $λ_{C u} / λ_{M o}$ is close to

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1.99

2.14

0.50

0.48

answer is B.

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Detailed Solution

$∵ \frac{1}{λ} = R (z - b^{2}) (\frac{1}{n_{i}^{2}} - \frac{1}{n_{i}^{2}})$ for K-series, b = 1 and for $k_{\propto}$

Transition $n_{f} = 2 t o n_{i} = 1 ∴ \frac{1}{λ} \propto {(z - 1)}^{2}$

$∵ \frac{λ_{C u}}{λ_{M o}} = \frac{{(Z_{M o} - 1)}^{2}}{{(Z_{C u} - 1)}^{2}} = {(\frac{42 - 1}{29 - 1})}^{2} = {(\frac{41}{28})}^{2} = 2.14$

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