Question

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

Answer 1

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,
${\lambda _\alpha } = 0.76\,A^\circ$
$\Rightarrow \,{\lambda _\alpha } = 0.76 \times {10^{ - 10}}$
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
$\dfrac{1}{{{\lambda _\alpha }}} = \,\dfrac{3}{4}R{(Z - 1)^2}$
Now, putting the values of ${\lambda _\alpha }$ and $R$ in the above formula, we get
$\dfrac{1}{{0.76 \times {{10}^{ - 10}}}} = \,\dfrac{3}{4} \times 1.097 \times {10^7} \times {(Z - 1)^2}$
$\Rightarrow {(Z - 1)^2} = \,\dfrac{4}{3} \times \dfrac{1}{{0.76 \times {{10}^{ - 10}} \times 1.097 \times {{10}^7}}}$
$\Rightarrow {(Z - 1)^2} \cong \,1600$
$\Rightarrow \,Z - 1 = 40$
$\Rightarrow Z = 41$
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
$\dfrac{1}{\lambda } = R{Z^2}(\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}})$
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
$\dfrac{1}{{{\lambda _\alpha }}} = R{(Z - 1)^2}[\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}]$
$\Rightarrow \dfrac{1}{{{\lambda _\alpha }}} = R{(Z - 1)^2}[1 - \dfrac{1}{4}]$
$\Rightarrow \dfrac{1}{{{\lambda _\alpha }}} = R{(Z - 1)^2}[\dfrac{{4 - 1}}{4}]$
$\Rightarrow \dfrac{1}{{{\lambda _\alpha }}} = \dfrac{3}{4}R{(Z - 1)^2}$
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 $45$.