Question

The wavelength of the ${k_a}$ for an element of atomic number 43 is $\lambda $then the wavelength of ${k_a}$for an element of atomic number 29 is-
a. $\left( {\dfrac{{43}}{{29}}} \right)\lambda $
b. $\left( {\dfrac{{42}}{{28}}} \right)\lambda $
c. $\left( {\dfrac{9}{4}} \right)\lambda $
d. $\left( {\dfrac{4}{9}} \right)\lambda $

Answer 1

Hint: The relation of wavelength of an element with its atomic number is given by Moseley’s law which is as follows,
$\lambda \propto \dfrac{1}{{{{\left( {z - 1} \right)}^2}}}$

Complete step by step answer:
From the Moseley’s law we have-
$\lambda \propto \dfrac{1}{{{{\left( {z - 1} \right)}^2}}}$ …….(1)
Where z=atomic number
So for the first case it is given that z=43
Therefore, $\lambda \propto \dfrac{1}{{{{\left( {43 - 1} \right)}^2}}}$ ……..(2)
Similarly for the second case z=29
Therefore, $\lambda ' \propto \dfrac{1}{{{{\left( {29 - 1} \right)}^2}}}$ ….(3)
Now divide equation (3) and (2) we get,
$ \Rightarrow \dfrac{{\lambda '}}{\lambda } = \dfrac{{{{\left( {43 - 1} \right)}^2}}}{{{{\left( {29 - 1} \right)}^2}}}$
$ \Rightarrow \dfrac{{\lambda '}}{\lambda } = \dfrac{{{{(42)}^2}}}{{{{(28)}^2}}}$
$ \Rightarrow \dfrac{{\lambda '}}{\lambda } = \dfrac{{{{(42)}^2}}}{{{{(28)}^2}}}$
$ \Rightarrow \dfrac{{\lambda '}}{\lambda } = \dfrac{9}{4}$
$ \Rightarrow \lambda ' = \left( {\dfrac{9}{4}} \right)\lambda $

Hence, the correct answer is option (B).

Note: Never apply the full equation of Moseley’s law in such questions. Look at the factors on which it depends and use the proportionate equation accordingly.