Question

An element of atomic number $9$ units ${{K}_{\alpha }}$ X-ray of wavelength $\lambda $. Find the atomic number of the element which emits ${{K}_{\alpha }}$ X-ray of wavelength $4\lambda $.

Answer 1

HintWe use the formula
     $\dfrac{1}{\lambda }=R{{\left( Z-1 \right)}^{2}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)$ …………………(1)
Where $\lambda =$ Wavelength, $Z=$ Atomic number, and $R=$ Rydberg constant

 Complete step-by-step solution:
$ K-alpha:- {{K}_{\alpha }}$ is typically by for the strongest X-Ray spectral lines for an element bombarded with energy sufficient to cause maximally intense X-Ray emission.
Moseley equation:- It states that the frequency of the spectral line in the characteristic X-Ray spectrum is directly proportional to the square of the atomic number of the element considered.
$\sqrt{f}=a\left( Z-b \right)$
According to the question,
$\begin{align}
  & {{Z}_{1}}=9 \\
 & {{\lambda }_{1}}=\lambda \\
\end{align}$
Then by the equation (1)
$\begin{align}
  & \dfrac{1}{\lambda }=R{{\left( 9-1 \right)}^{2}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right).............(2) \\
 & {{Z}_{2}}=Z \\
 & {{\lambda }_{2}}=4\lambda \\
\end{align}$
$\dfrac{1}{4\lambda }=R{{\left( Z-1 \right)}^{2}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)...............(3)$
Equation (2) is divided by equation (3)
$\dfrac{\dfrac{1}{\lambda }}{\dfrac{1}{4\lambda }}=\dfrac{R\left( 9-1 \right)}{R\left( Z-1 \right)}\dfrac{\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)}{\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)}$
$4=\dfrac{{{\left( 8 \right)}^{2}}}{{{\left( Z-1 \right)}^{2}}}$
$\begin{align}
  & {{\left( Z-1 \right)}^{2}}=\dfrac{8\times 8}{4} \\
 & {{\left( Z-1 \right)}^{2}}=16 \\
\end{align}$
$\begin{align}
  & Z-1=4 \\
 & Z=4+1 \\
 & Z=5 \\
\end{align}$

Note: Student think that when the wavelength are different, then the transition are also different but transition takes place for both between same states.