Question

What element has ${K_\alpha }$ line of wavelength $1.785{A^o}$ ? $R = 109737c{m^{ - 1}}$
(A) Platinum
(B) Zinc
(C) Iron
(D) Cobalt

Answer 1

First we need to understand the concept of wavelength to solve this question. Wavelength is the distance between corresponding points of two consecutive waves. Corresponding points refers to two points or particles in the same phase.

Complete answer:
The atomic number or proton number (which is represented by $Z$ ) of a chemical element is the number of protons present in the nucleus of every atom of that particular element. The atomic number uniquely identifies a chemical element. It is identical to the charge number of the nucleus.
When an electron vacancy in the K shell is filled by an electron from the L shell, the characteristic energy/wavelength of the emitted photon is called the ${K_\alpha }$ spectral line
The wavelength of a ${K_\alpha }$ line is related to the atomic number $Z$ by Moseley’s formula, which is
$\dfrac{1}{{{\lambda _{{k_\alpha }}}}} = R{(Z - 1)^2}\left( {\dfrac{1}{{n_1^2 - n_2^2}}} \right)$
We know that for a ${K_\alpha }$ line,
$\dfrac{1}{{{\lambda _{{k_\alpha }}}}} = R{(Z - 1)^2}\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right)$
$\dfrac{1}{{{\lambda _{{k_\alpha }}}}} = R{(Z - 1)^2}\left( {\dfrac{1}{1} - \dfrac{1}{4}} \right)$
$\dfrac{1}{{{\lambda _{{k_\alpha }}}}} = R{(Z - 1)^2}\left( {\dfrac{3}{4}} \right)$
On taking ${(Z - 1)^2}$ on one side and all the other terms on the other side, we get,
${(Z - 1)^2} = \dfrac{4}{3}\dfrac{1}{{{\lambda _{{k_\alpha }}}}}\dfrac{1}{R}$
On putting the required values,
${(Z - 1)^2} = \dfrac{4}{3} \times \dfrac{1}{{1.785 \times {{10}^{ - 8}}}} \times \dfrac{1}{{109737}}$
${(Z - 1)^2} = 680.6$
On taking square root on both the sides,
$Z - 1 = 26$
$Z = 27$
We know that the atomic number of cobalt is $27$ .
So, the correct answer is (D) Cobalt.

Note:
It is important to note that the elements having higher atomic number gives higher energy waves which are of shorter wavelength. When an electron vacancy in the K shell is filled by an electron from the L shell, the characteristic energy/wavelength of the emitted photon is called the ${K_\alpha }$ spectral line.