Question

The wavelength of ${{K}_{\alpha }}$ X-rays for lead isotopes $P{{b}^{208}}$, $P{{b}^{206}}$ and $P{{b}^{204}}$ are ${{\lambda }_{1}}$, ${{\lambda }_{2}}$ and ${{\lambda }_{3}}$ respectively. Then
(A) ${{\lambda }_{2}}=\sqrt{{{\lambda }_{1}}{{\lambda }_{3}}}$
(B) ${{\lambda }_{2}}={{\lambda }_{1}}+{{\lambda }_{3}}$
(C) ${{\lambda }_{2}}={{\lambda }_{1}}{{\lambda }_{3}}$
(D) ${{\lambda }_{2}}=\dfrac{{{\lambda }_{1}}}{{{\lambda }_{3}}}$

Answer 1

In the given question, we have been provided with information on three isotopes of lead and we have been asked a relation between the wavelengths of their respective ${{K}_{\alpha }}$ X-rays. A ${{K}_{\alpha }}$ X-ray is produced when the electron in an element transitions to the first energy level or the first shell of the atom from a comparatively higher energy level. Let’s see in detail how we can relate the given wavelengths.

Formula Used $\sqrt{\upsilon }=k.Z$, $c=\upsilon \times \lambda$

Complete step by step answer:
From Moseley’s law, we know that the square root of the frequency of the characteristic X-ray depends on the atomic number of the element.
We can represent Moseley’s law as $\sqrt{\upsilon }=k.Z$ where $\upsilon$ is the frequency of the X-rays, $k$ is the constant of proportionality and $Z$ is the atomic number of the element.
For electromagnetic radiations, we know that $c=\upsilon \times \lambda$ where $c$ is the speed of the light and has a constant value and $\lambda$ is the wavelength of the radiation.
From the two equations given above, we can say that $\sqrt{\dfrac{c}{\lambda }}=k.Z$ where the meanings of the symbols have been discussed above
The wavelength and the atomic number are the only variables in the above equation and the other quantities are constant, hence we can say that the variation of the wavelength of the characteristic X-rays depends on the atomic number of the element.
Since isotopes of an element have the same atomic number, we can say that all isotopes of an element will possess the same wavelength for the characteristic X-rays.
Therefore, ${{\lambda }_{2}}={{\lambda }_{1}}={{\lambda }_{3}}$ is the correct relation for the wavelengths given.

When we try to substitute the above relation in the given options, we will find that option (A) is the correct answer.

Note
We could directly say that option (A) is the correct answer because option (A) represents that ${{\lambda }_{1}}$ is the geometric mean of the wavelengths ${{\lambda }_{2}}$ and ${{\lambda }_{3}}$. Now we all know that every number is a geometric mean of itself, hence it will satisfy the equality condition obtained in the solution.
You should be aware of the basic meaning of the terms such as isotopes, isobars, isoelectronic species as your solution depends on your interpretation of the question.