Advanced Modern Physics
2.0 Moseley's Law.
2.0 Moseley's Law.
Moseley measured and plotted the square root of frequency ($v$) of characteristics X-rays of a large number of elements against their atomic number ($Z$).
The plot was found to be a nearly straight line not passing through the origin. He gave the equation of the straight line as,
$$\sqrt v = a\left( {Z - b} \right)$$
where $a$ and $b$ are positive constants for one type of X-ray and all elements. This is known as Moseley's law.
It can be derived from the Bohr's theory of the atom.
Frequency of the characteristic X-ray is given by,
$$\sqrt v = \sqrt {RC\left( {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right)} \left( {Z - b} \right){\text{ (for multi electron system) }}$$
where,
$b$ is known as screening constant,
$\left( {Z - b} \right)$ is the effective nuclear charge.
Example:
For ${K_\alpha }$, line $n_1$=1 and $n_2$=2, thus
$$\begin{equation} \begin{aligned} \sqrt v = \sqrt {\frac{{3RC}}{4}} \left( {Z - b} \right) \\ \\\end{aligned} \end{equation} $$ So, $$a = \sqrt {\frac{{3RC}}{4}} \quad \left( {b = 1,\;{\text{for}}\;{K_\alpha }{\text{ line}}} \right)$$
Question 4. Which of the material of $A$ and $B$ have the highest atomic number in $Fig. 6$?
Solution:
As constants $a$ and $b$ are same for a particular type of characteristic X-rays, so we take into account only one type of characteristic X-ray of both material (say ${K_\alpha }$ ).
$$\sqrt v = \sqrt {RC\left( {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right)} \left( {Z - b} \right)$$ So, $$\begin{equation} \begin{aligned} \Rightarrow {\lambda _{{K_{{\alpha _{(A)}}}}}} < {\lambda _{{K_{{\alpha _{(B)}}}}}} \\ \Rightarrow {v_{{K_{{\alpha _{(A)}}}}}} > {v_{{K_{{\alpha _{(B)}}}}}}\quad \; \\ \Rightarrow {Z_A} > {Z_B} \\\end{aligned} \end{equation} $$
Question 5. A cobalt target ($Z$=27) is bombarded with electrons and the wavelength of its characteristic spectrum are measured.
A second, fainter, characteristic spectrum is also found because of an impurity in the target. The wavelength of the ${K_\alpha }$ lines are 178.9 $pm$ (cobalt) and 143.5 $pm$ (impurity). What is the impurity?
Solution:
Using Moseley's law and putting $\left( {\frac{c}{\lambda }} \right)$ for $v$ (and assuming $b=1$), we obtain
$$\sqrt {\frac{c}{{{\lambda _{Co}}}}} = a{Z_{Co}} - a{\text{ }}$$ and $$\sqrt {\frac{c}{{{\lambda _x}}}} = a{Z_x} - a$$ Dividing the above two equation we get, $$\begin{equation} \begin{aligned} \sqrt {\frac{{{\lambda _{Co}}}}{{{\lambda _x}}}} = \frac{{{Z_x} - 1}}{{{Z_{Co}} - 1}} \\ \sqrt {\frac{{178.9{\text{ pm}}}}{{143.5{\text{ pm}}}}} = \frac{{{Z_x} - 1}}{{27 - 1}} \\\end{aligned} \end{equation} $$ After solving, $${Z_x} = 30.0$$
k
So, the impurity is zinc $(Zn)$.
