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,the wavelength $\lambda$ is given by: $\frac{1}{\lambda} = R(Z-1)^2 \left( \frac{1}{1^2} - \f

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$4$

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(D) According to Moseley's law for $K_{\alpha}$ $X$-rays,the wavelength $\lambda$ is given by: $\frac{1}{\lambda} = R(Z-1)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R(Z-1)^2 \left( \frac{3}{4} ight)$.For metal $A$: $\frac{1875R}{4} = R(Z_A-1)^2 \left( \frac{3}{4} ight) \Rightarrow (Z_A-1)^2 = \frac{1875}{4} 	imes \frac{4}{3} = 625 \Rightarrow Z_A-1 = 25 \Rightarrow Z_A = 26$.For metal $B$: $675R = R(Z_B-1)^2 \left( \frac{3}{4} ight) \Rightarrow (Z_B-1)^2 = 675 	imes \frac{4}{3} = 900 \Rightarrow Z_B-1 = 30 \Rightarrow Z_B = 31$.The number of elements between $A$ and $B$ is given by $|Z_B - Z_A| - 1 = |31 - 26| - 1 = 5 - 1 = 4$.

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 the Rydberg constant. The number of elements lying between $A$ and $B$ according to their atomic numbers is:

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