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Question: The wavelength of the \({k_a}\) for an element of atomic number 43 is \(\lambda \)then the wavelengt...

The wavelength of the ka{k_a} for an element of atomic number 43 is λ\lambda then the wavelength of ka{k_a}for an element of atomic number 29 is-
a. (4329)λ\left( {\dfrac{{43}}{{29}}} \right)\lambda
b. (4228)λ\left( {\dfrac{{42}}{{28}}} \right)\lambda
c. (94)λ\left( {\dfrac{9}{4}} \right)\lambda
d. (49)λ\left( {\dfrac{4}{9}} \right)\lambda

Explanation

Solution

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

Complete step by step answer:
From the Moseley’s law we have-
λ∝1(z−1)2\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, λ∝1(43−1)2\lambda \propto \dfrac{1}{{{{\left( {43 - 1} \right)}^2}}} ……..(2)
Similarly for the second case z=29
Therefore, λ′∝1(29−1)2\lambda ' \propto \dfrac{1}{{{{\left( {29 - 1} \right)}^2}}} ….(3)
Now divide equation (3) and (2) we get,
⇒λ′λ=(43−1)2(29−1)2\Rightarrow \dfrac{{\lambda '}}{\lambda } = \dfrac{{{{\left( {43 - 1} \right)}^2}}}{{{{\left( {29 - 1} \right)}^2}}}
⇒λ′λ=(42)2(28)2\Rightarrow \dfrac{{\lambda '}}{\lambda } = \dfrac{{{{(42)}^2}}}{{{{(28)}^2}}}
⇒λ′λ=(42)2(28)2\Rightarrow \dfrac{{\lambda '}}{\lambda } = \dfrac{{{{(42)}^2}}}{{{{(28)}^2}}}
⇒λ′λ=94\Rightarrow \dfrac{{\lambda '}}{\lambda } = \dfrac{9}{4}
⇒λ′=(94)λ\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.