Question

The wavelength of the ${{{K}}_{{\alpha }}}$line for an element of atomic number $57$is${{\lambda }}$. What is the wavelength of ${{{K}}_{{\alpha }}}$ line for the element of atomic number$29$?
A. $\lambda$
B. $2\lambda$
C. $4\lambda$
D. $\dfrac{{{\lambda }}}{{{4}}}$

Answer 1

The relation between wavelength of an element and the atomic number of an element is given by the formula, $\dfrac{{{1}}}{{{\lambda }}}{{ \alpha (z - 1}}{{{)}}^{{2}}}$. This formula can also be written as $\dfrac{{{{{\lambda }}_{{2}}}}}{{{{{\lambda }}_{{1}}}}}{{ = }}\dfrac{{{{{{(}}{{{z}}_{{1}}}{{ - 1)}}}^{{2}}}}}{{{{{{(}}{{{z}}_{{2}}}{{ - 2)}}}^{{2}}}}}$. Then substitute the given values of atomic number and wavelength in the formula. Now, simply it by doing further calculations and then the required relation can be obtained.

Complete step by step answer:
The atomic number or proton number (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 (similar) to the charge number of the nucleus.

Given: Atomic number of an element, ${{{z}}_1}{{ = 57}}$
Wavelength of ${{{K}}_{{\alpha }}}$line of an element is ${{\lambda }}$
To find: Value of wavelength ${{{K}}_{{\alpha }}}$line for element having atomic number, ${{{z}}_{{2}}}{{ = 29}}$

Let the wavelength of an element having atomic number $29$ be ${{\lambda '}}$.
The relationship between wavelength and atomic number for ${{{K}}_{{\alpha }}}$line is given by
$\dfrac{{{1}}}{{{\lambda }}}{{ \alpha (z - 1}}{{{)}}^{{2}}} \\\ \text{ or} \dfrac{{{{\lambda '}}}}{{{\lambda }}}{{ = }}\dfrac{{{{{{(}}{{{z}}_{{1}}}{{ - 1)}}}^{{2}}}}}{{{{{{(}}{{{z}}_{{2}}}{{ - 1)}}}^{{2}}}}} \\\$

On substituting the values in above formula, we get
$\dfrac{\lambda '}{\lambda}=\dfrac{(57-1)^2}{(29-1)^2}=\dfrac{4}{1}$
${\lambda '}=4\lambda$

Thus, the wavelength of ${{{K}}_{{\alpha }}}$ line for the element having atomic number $29$ is four times that of the wavelength of an element having atomic number $57$is${{4\lambda }}$.

Hence, the correct answer is option (C).

Additional information:

Note: The wavelength of ${{{K}}_{{\alpha }}}$ line is related to atomic number ${{Z}}$ by Moseley's Formula $\dfrac{{{1}}}{{{\lambda }}}{{ = R(Z - 1}}{{{)}}^{{2}}}\left( {\dfrac{{{1}}}{{{{{n}}_{{1}}}^{{2}}}}{{ - }}\dfrac{{{1}}}{{{{{n}}_{{2}}}^{{2}}}}} \right)$
The elements having high atomic number (for example molybdenum) give high energy rays (with short wavelengths).
K-alpha emission lines are in actuality the result when an electron transitions to the innermost "K" shell (having principal quantum number 1) from a 2p orbital of the second or "L" shell (having principal quantum number 2). The K-alpha1 emission (represented by ${{{k}}_{{{{\alpha }}_{{1}}}}}$) is higher in energy and therefore has a lower wavelength than the K-alpha2 emission (represented by ${{{k}}_{{{{\alpha }}_2}}}$).