Question

The wavelength of \[{K_\alpha }\] X-rays of two metals A and B are \[\dfrac{4}{{1875R}}\] and \[\dfrac{1}{{675R}}\], respectively, where ’R’ is Rydberg's constant. The number of elements lying between A and B according to their atomic number is
A. 3
B. 6
C. 5
D. 4

Answer 1

Hint:In this question we use the concept of Moseley’s law. Here we will use the formula of Moseley’s law for the wavelength of \[{K_\alpha }\] X-rays. The Rydberg formula is used to determine the wavelength of the light emitted by an electron which is moving between the energy levels of an atom.

Formula used:
The Rydberg equation for hydrogen is given as,
\[\dfrac{1}{\lambda } = R{(Z - 1)^2}\left[ {\dfrac{1}{{{n_2}^2 - {n_1}^2}}} \right]\]
Where \[\lambda \] is the wavelength of light.
R is the Rydberg constant.
Z is the atomic number.
\[{n_1}\]and \[{n_2}\] is the principal quantum number of the upper and lower energy levels.

Complete step by step solution:
Given the wavelength of two metals A and B are,
\[{\lambda _A} = \dfrac{4}{{1875R}}\] and \[{\lambda _B} = \dfrac{1}{{675R}}\]
As we know that wavelength,
\[\dfrac{1}{\lambda } = R{(Z - 1)^2}\left[ {\dfrac{1}{{{n_2}^2 - {n_1}^2}}} \right] \\ \]
The wavelength \[\lambda \] of \[{K_\alpha }\] X-rays originates from \[{n_2} = 2\] to \[{n_1} = 1\].
For metal A,
\[\dfrac{{1875R}}{4} = R{({Z_1} - 1)^2}\left[ {\dfrac{3}{4}} \right]\]

By solving this we get
\[{Z_1} = 26\]
Similarly for metal B,
\[\dfrac{{675R}}{1} = R{({Z_2} - 1)^2}\left[ {\dfrac{3}{4}} \right] \\ \]
By solving this we get
\[{Z_1} = 31\]
Now \[{Z_2} - {Z_1} = 31 - 26 = 4\]
Therefore, the number of elements lying between A and B according to their atomic number is 4.

Hence option D is the correct answer.

Note:When the vacancy of an electron in the K shell is filled by an electron from the L shell then the characteristic energy/wavelength of the emitted photon is known as the K-alpha (\[{K_\alpha }\]) spectral line. When an electron transfers from one atomic orbital to another then there must be a change in energy.