Question

If ${\lambda _{cu}}$ is the wavelength of ${K_\alpha }$, X-ray line of copper (atomic number 29) and ${\lambda _{Mo}}$
​ is the wavelength of the ${K_\alpha }$ X-ray line of molybdenum (atomic number 42), then the ratio ${\lambda _{cu}}/{\lambda _{mo}}$
is close to :
A. 1.99
B. 2.14
C. 0.50
D. 0.48

Answer 1

The above problem can be resolved using the m=basic mathematical relation for the wavelength and the atomic number of any element. This relation is enlisted for both the elements given in the problem, namely the copper and the molybdenum. Moreover, this relation signifies that the wavelength persists some relation with the atomic number of the elements. Moreover, this analysis can be taken over for the various elements of the definite value of their wavelengths or the atomic numbers, belonging to the specific class of elements, where the wavelength and the atomic numbers possess its own significance in the analysis.

Complete step by step answer:
Given:
The atomic number of copper is, ${Z_1} = 29$.
The atomic number of molybdenum is, ${Z_2} = 42$
The mathematical relation between the wavelength and the atomic number is,
$\lambda = {\left( {Z - 1} \right)^{ - 2}}$
Taking the ratio for the wavelength of copper and molybdenum as,

\dfrac{{{\lambda _{cu}}}}{{{\lambda _{Mo}}}} = \dfrac{{{{\left( {{Z_1} - 1} \right)}^{ - 2}}}}{{{{\left( {{Z_2} - 1} \right)}^{ - 2}}}}\\\ \dfrac{{{\lambda _{cu}}}}{{{\lambda _{Mo}}}} = {\left( {\dfrac{{{Z_2} - 1}}{{{Z_1} - 1}}} \right)^2}\\\ \dfrac{{{\lambda _{cu}}}}{{{\lambda _{Mo}}}} = {\left( {\dfrac{{42 - 1}}{{29 - 1}}} \right)^2}\\\ \dfrac{{{\lambda _{cu}}}}{{{\lambda _{Mo}}}} = 2.144 \end{array}$$ Therefore, the ratio $${\lambda _{cu}}/{\lambda _{mo}}$$ is close to 2.14 **So, the correct answer is “Option B”.** **Note:** Try to identify the mathematical formulation for the wavelengths of the given element and its relation with the desired atomic number. The appropriate relation has its significance in various applications like determining the hydrogen series like Lyman series, Balmer series, Paschen series and many more. Moreover, this relation needs to be taken care of the correct evaluation of the atomic numbers along with classifications in accordance with the period; they belong to a modern periodic table. The Wavelength of any particular element has its significance in finding the frequency along with the other parameters, undertaken in specific experimentation.