Question

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

• A. $3$
• B. $1$
• C. $4$
• D. $5$

Answer 1

The correct option is C $4$

Using $\frac{1}{\lambda }=R(Z-1{)}^2[\frac{1}{n_2^2}-\frac{1}{n_1^2}]$
For $\alpha$ particle, $n_1=2,n_2=1$
For metal $A:\frac{1875R}{4}=R(Z_1-1{)}^2\left(\frac{3}{4}\right)\Rightarrow Z_1=26$
For metal $B:675R=R(Z_2-1{)}^2\left(\frac{3}{4}\right)\Rightarrow Z_2=31$
Therefore, 4 elements lie between A and B.