Question

The element which has a ${K_\alpha }$​ X-rays line of wavelength $1.8\mathop A\limits^o $ is ? $(R = 1.1 \times {10^7}{m^{ - 1}},\,b = 1,\,\sqrt {\dfrac{5}{{33}}} = 0.39)$
A. $Co,\,Z = 27$
B. Iron, $Z = 26$
C. $Mn,\,Z = 25$
D. $Ni,\,Z = 28$

Answer 1

Hint:In order to answer this question, we must first comprehend the concept of wavelength. The wavelength is the separation between the matching spots of two subsequent waves. Two points or particles that are in the same phase are said to be corresponding points.

Formula used:
The formula of Moseley which relates the wavelength of the spectral line is given by:
$\dfrac{1}{{{\lambda _{{K_\alpha }}}}} = R{(Z - 1)^2}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$
Where, $n_1, n_2$ are the number of orbits, $Z$ is the atomic number, $R$ is the Rydberg constant and $\lambda$ is the wavelength.

Complete step by step solution:
In the question, we have given that the wavelength of ${K_\alpha }$ spectral line is $1.8\mathop A\limits^o $ and $R = 1.1 \times {10^7}{m^{ - 1}},\,b = 1,\,\sqrt {\dfrac{5}{{33}}} = 0.39$. In order to know that a chemical element's atomic number, also known as its proton number $(Z)$, refers to the number of protons found in each atom's nucleus. The atomic number is a distinctive way to identify a chemical element and the charge number of the nucleus is the same.

The characteristic energy or wavelength of the produced photon when an electron from the $L$ shell fills an electron vacancy in the $K$ shell is known as the ${K_\alpha }$ spectral line. Moseley's formula, which is, relates the wavelength of a ${K_\alpha }$ line to the atomic number $Z$:
$\dfrac{1}{{{\lambda _{{K_\alpha }}}}} = R{(Z - 1)^2}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right) \\ $
For ${K_\alpha }$line, substitute the given values in the above formula, then we have:
$\dfrac{1}{{1.8 \times {{10}^{ - 12}}}} = (1.1 \times {10^7}){(Z - 1)^2}\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right) \\$
$\Rightarrow \dfrac{1}{{1.8 \times {{10}^{ - 12}}}} = (1.1 \times {10^7}){(Z - 1)^2}\left( {\dfrac{3}{4}} \right) \\$
$\Rightarrow Z - 1 = 26 \\$
$\Rightarrow Z = 27 \\$
Thus, the correct option is A.

Note: It should be noted that the high-atomic-number elements emit high-energy rays. In reality, the formation of ${K_\alpha }$ emission lines happens when an electron moves from a $2p$ orbital in the second or $L$ shell to the innermost $K$ shell, which has the main quantum number $1$. Since the ${K_{{\alpha _2}}}$ emission has a higher energy than the ${K_{{\alpha _2}}}$ emission, it has a shorter wavelength.