The wavelengths of K_{alpha} X-rays of two me | vedclass

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(D) According to Moseley's law for $K_{\alpha}$ $X$-rays: $\frac{1}{\lambda} = R(Z-1)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = \frac{3R}{4}(Z-

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$4$

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(D) According to Moseley's law for $K_{\alpha}$ $X$-rays: $\frac{1}{\lambda} = R(Z-1)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = \frac{3R}{4}(Z-1)^2$.For metal $A$: $\frac{1875R}{4} = \frac{3R}{4}(Z_A-1)^2 \implies (Z_A-1)^2 = 625 \implies Z_A-1 = 25 \implies Z_A = 26$.For metal $B$: $675R = \frac{3R}{4}(Z_B-1)^2 \implies (Z_B-1)^2 = 900 \implies Z_B-1 = 30 \implies Z_B = 31$.The number of elements between $A$ and $B$ is given by $Z_B - Z_A - 1 = 31 - 26 - 1 = 4$.

The wavelengths of $K_{\alpha}$ $X$-rays of two metals $A$ and $B$ are $\frac{4}{1875R}$ and $\frac{1}{675R}$ respectively,where $R$ is Rydberg's constant. The number of elements lying between $A$ and $B$ according to their atomic numbers is

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