Question

The atomic number $\left(Z\right)$ of an element whose $k_{\alpha }$ wavelength is $\lambda$ is $11$. The atomic number of an element whose $k_{\alpha }$ wavelength is $4\lambda$ is equal to

• A. $6$
• B. $11$
• C. $44$
• D. $4$

Answer 1

The correct option is A $6$

$\begin{array}{c}\text{we know that for}{k}_{\alpha }\text{spectral line-}\\ \begin{array}{cccc}& \frac{1}{{\lambda }_{\alpha }}\alpha (z-1{)}^2& \text{where}{\lambda }_{\alpha }=\text{wavelength}& \\ & z=\text{atomic number}& & \\ \Rightarrow & \frac{1}{{\lambda }_{\alpha }}=k\cdot (z-1{)}^2& & \end{array}\end{array}$

$\begin{array}{c}\text{For case}1:\text{we have}{\lambda }_{\alpha }=\lambda \text{and}z=11\\ \Rightarrow \frac{1}{\lambda }=k\cdot (10{)}^2\Rightarrow k=\frac{1}{100}\lambda \\ \text{For}\mathrm{case}2\to \text{we have}{\lambda }_{\alpha }=4\lambda \\ \Rightarrow \frac{1}{4\lambda }=k\cdot (z-1{)}^2=\frac{1}{100}\lambda \cdot (z-1{)}^2\end{array}$

$\begin{array}{c}\Rightarrow (z-1{)}^2=25\\ z-1=5\\ \Rightarrow z=6\end{array}$