Question

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

• A. $3$
• B. $6$
• C. $5$
• D. $4$

Answer 1

The correct option is D $4$

In X-ray spectroscopy, K-alpha emission lines result when an electron transitions to the innermost "K" shell (principal quantum number 1) from a 2p orbital of the second or "L" shell (with principal quantum number 2).
$\frac{1}{\lambda }=Z^{2}R(1-\frac{1}{2^2})=\frac{3R{Z}^2}{4}$
${\lambda }_A=\frac{4}{1875R}=\frac{4}{3R{Z}^2}$
$Z_A=25$
Similarly,
${\lambda }_B=\frac{1}{675R}=\frac{4}{3R{Z}_B^2}$
$Z_B=30$

No. of elements between A and B $=4$