$(a)\;1.5A^{0}\qquad(b)\;2.6A^{0}\qquad(c)\;3.6A^{0}\qquad(d)\;4.2A^{0}$

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Answer : (a) $\;1.5A^{0}$

Explanation :

According to Moseley's low

$\sqrt{\nu}=a(Z-1)$

$(Z-1)^2\; \propto \nu \quad \; or \quad \; (Z-1)^2 \propto \large\frac{1}{\lambda}$

$\large\frac{(Z_{M_{0}-1})^{2}}{(Z_{Cu}-1)^2}=\large\frac{\lambda_{Cu}}{\lambda_{M_{0}}}$

$\lambda_{Cu}=\lambda_{M_{0}}\;\large\frac{(Z_{M_{0}-1})^2}{(Z_{cu}-1)^2}$

$=0.71\times(\large\frac{41}{28})^2\;A^{0}$

$=1.52 A^{0}\;.$

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