Question

The wavelength of ${K_\alpha }$ line for an element of atomic number 43 is ‘$\lambda$’. Then the wavelength of ${K_\alpha }$ line for an element of atomic number 29 is
A) $\dfrac{9}{4}\lambda$.
B) $\dfrac{{43}}{{29}}\lambda$.
C) $\dfrac{4}{9}\lambda$.
D) $\dfrac{{42}}{{28}}\lambda$.

Answer 1

The formula of Moseley’s law for k alpha line can be used for the calculation for the correct answer for this problem. The formula for Moseley’s law for the k alpha line contains constants a and b, which has the value b=1. The value of constants a and b depends on the series.

Formula: The formula for Moseley’s law for k alpha line is $\lambda \propto \dfrac{1}{{{{\left( {Z - b} \right)}^2}}}$ where $\lambda$ is the wavelength, $Z$ is the atomic number and $b$ is constant having value of $b = 1$ for k alpha line.

Step by step solution:
Step 1.
As the Moseley’s law for alpha line is $\lambda \propto \dfrac{1}{{{{\left( {Z - b} \right)}^2}}}$ where $\lambda$ is the wavelength, $Z$ is the atomic number and $b$ is constant having value of $b = 1$ for k alpha line.
For atomic number 43
$\lambda \propto \dfrac{1}{{{{\left( {Z - b} \right)}^2}}} \\\ {\lambda _{43}}\propto \dfrac{1}{{{{\left( {43 - 1} \right)}^2}}} \\\ {\lambda _{43}}\propto \dfrac{1}{{{{\left( {42} \right)}^2}}} \\\$………eq.(1)
Step 2.
As the Moseley’s law for alpha line is $\lambda \propto \dfrac{1}{{{{\left( {Z - b} \right)}^2}}}$ where $\lambda$ is the wavelength, $Z$ is the atomic number and $b$ is constant having value of $b = 1$ for k alpha line.
For atomic number 29
$\lambda \propto \dfrac{1}{{{{\left( {Z - b} \right)}^2}}} \\\ {\lambda _{29}}\propto \dfrac{1}{{{{\left( {29 - 1} \right)}^2}}} \\\ {\lambda _{29}}\propto \dfrac{1}{{{{\left( {28} \right)}^2}}} \\\$………eq.(2)
Step 3.
We can get the value of ${\lambda _{29}}$ by dividing equation 2nd by equation 1st
Therefore applying$\dfrac{{\left( 2 \right)}}{{\left( 1 \right)}}$,
$\dfrac{{{\lambda _{29}}}}{{{\lambda _{43}}}} = \dfrac{{{{\left( {42} \right)}^2}}}{{{{\left( {28} \right)}^2}}} \\\ {\lambda _{29}} = {\left( {\dfrac{{42}}{{28}}} \right)^2} \cdot {\lambda _{43}} \\\$………eq.(3)
As the wavelength of atomic number 43 is $\lambda$
So,
${\lambda _{43}} = \lambda$
Replacing ${\lambda _{43}} = \lambda$ in equation (3).
${\lambda _{29}} = {\left( {\dfrac{{42}}{{28}}} \right)^2} \cdot {\lambda _{43}} \\\ {\lambda _{29}} = {\left( {\dfrac{{42}}{{28}}} \right)^2} \cdot \lambda \\\$………eq.(4)
Step 4.
Solving the equation (4) we get
${\lambda _{29}} = {\left( {\dfrac{{42}}{{28}}} \right)^2} \cdot \lambda \\\ {\lambda _{29}} = \dfrac{9}{4} \cdot \lambda \\\$
So the correct answer for this problem is ${\lambda _{29}} = \dfrac{9}{4} \cdot \lambda$ i.e. option A.

Option A is correct

Note: While solving this question students should know that Moseley’s law says that $\lambda = \dfrac{c}{{{a^2}{{\left( {Z - b} \right)}^2}}}$ where $c$ is speed of light $a$ and $b$ are constants and $Z$ is the atomic number the value of $b$ is $b = 1$ for k alpha line. Also the values of $a$ and $b$ are independent of the material but depends on the X-ray series.