Question

The Wavelength $K_{\alpha }$ $X-$ rays produced by a $X-$ ray tube is $0.76A.$ the atomic number of anticathode material is:
A) $82$
B) $41$
C) $20$
D) $50$

Answer 1

Hint: This is the concept of Moseley’s law. Here will use the formula of Moseley’s law for calculating the wavelength of $K_{\alpha }X-rays$. Now, you might use the Rydberg equation here, but remember, the Rydberg equation is used to calculate the wavelengths of hydrogenic atoms. It cannot give the correct answer for $K_{\alpha }X-rays$.

Complete step by step solution:
In the question, we are given,
${\lambda }_{\alpha }=0.76A^{\circ }$
$\Rightarrow {\lambda }_{\alpha }=0.76\times {10}^{-10}$
Which is the wavelength of $K_{\alpha }X-$rays.
Also, $R=1.097\times {10}^{7}m$
Which is the value of the Rydberg constant.
Now, using the Rydberg equation in terms of Moseley’s law, which is given by
${\displaystyle \frac{1}{{\lambda }_{\alpha }}}={\displaystyle \frac{3}{4}}R(Z-1{)}^2$
Now, putting the values of ${\lambda }_{\alpha }$ and $R$ in the above formula, we get
${\displaystyle \frac{1}{0.76\times {10}^{-10}}}={\displaystyle \frac{3}{4}}\times 1.097\times {10}^7\times (Z-1{)}^2$
$\Rightarrow (Z-1{)}^2={\displaystyle \frac{4}{3}}\times {\displaystyle \frac{1}{0.76\times {10}^{-10}\times 1.097\times {10}^7}}$
$\Rightarrow (Z-1{)}^2\cong 1600$
$\Rightarrow Z-1=40$
$\Rightarrow Z=41$
Therefore, the atomic number of anticathode material is $41$.

Therefore, option (B) is the correct option.

Additional information:
As we all know, the Rydberg equation for hydrogen-like chemical elements is given by
${\displaystyle \frac{1}{\lambda }}=R{Z}^2({\displaystyle \frac{1}{{n_1}^2}}-{\displaystyle \frac{1}{{n_2}^2}})$
Where $\lambda$ is the wavelength of light is emitted, $R$ is the Rydberg constant, $Z$ is the atomic number, $n_1$ is the principal quantum number of upper energy level, and $n_2$ is the principal quantum number of lower energy level.
This equation is applicable to hydro genic atoms of chemical elements, but it is not applicable in case of $K_{\alpha }X-$ rays. Therefore, to calculate the wavelength in terms of $K_{\alpha }X-rays$, Moseley modifies the Rydberg equation by replacing $Z$ by $Z-1$ and putting $n_1=1$ and $n=2$, which is shown below
${\displaystyle \frac{1}{{\lambda }_{\alpha }}}=R(Z-1{)}^2[{\displaystyle \frac{1}{1^2}}-{\displaystyle \frac{1}{2^2}}]$
$\Rightarrow {\displaystyle \frac{1}{{\lambda }_{\alpha }}}=R(Z-1{)}^2[1-{\displaystyle \frac{1}{4}}]$
$\Rightarrow {\displaystyle \frac{1}{{\lambda }_{\alpha }}}=R(Z-1{)}^2[{\displaystyle \frac{4-1}{4}}]$
$\Rightarrow {\displaystyle \frac{1}{{\lambda }_{\alpha }}}={\displaystyle \frac{3}{4}}R(Z-1{)}^2$
Which is the equation for calculating wavelength in terms of $K_{\alpha }$ $X-rays$.

Note: Moseley used Moseley’s law to arrange $K$, $Ar$, $Co$ and $Ni$ in Mendeleev’s periodic table, where $K$ is the potassium, $Ar$ is Argon, $Co$ is the cobalt and $Ni$ is the nickel. Also, this law helped in the discovery of many elements like $Tc(43)$, $Pr(61)$ and $Rh(45)$, where $Tc(43)$ is the technetium with the atomic number $43$, $Pr(61)$ is the promethium with the atomic number $61$ and $Rh(45)$ is the rhodium with the atomic number $45$.