Question

In Coolidge tube if $f$ and $\lambda$ represents the frequency and wavelength respectively of $K_{\alpha }$ line for a metal of atomic number $Z$, then identify the option(s) which represents a straight line.

• A. $\sqrt{f}$ versus $Z$
• B. $\frac{1}{\sqrt{\lambda }}$ versus $Z$
• C. $f$ versus $Z$
• D. $\lambda$ versus $Z$

Answer 1

Answer: (A), (B)

$\frac{1}{\sqrt{\lambda }}$ versus $Z$

We know that, Moseley's law is given by,

$\sqrt{f}=a(Z-b)$

For $K_{\alpha }$ line, $b=1$

So, $\sqrt{f}=a(Z-1)$

$\Rightarrow \sqrt{f}=aZ-a...\left(1\right)$

This equation is similar to the equation of straight line, $y=mx+c$, where,
$y=\sqrt{f}$, $m=a$, $x=Z$, and $c=-a$

Hence, $\sqrt{f}$ versus $Z$ is a straight line.

From $\left(1\right)$,

$\sqrt{\frac{c}{\lambda }}=aZ-a$

$\frac{1}{\sqrt{\lambda }}=\frac{a}{\sqrt{c}}Z-\frac{a}{\sqrt{c}}$

This equation is also similar to the equation of straight line, $y=mx+C$, where,
$y=\frac{1}{\sqrt{\lambda }}$, $m=\frac{a}{\sqrt{c}}$, $x=Z$, and $C=-\frac{a}{\sqrt{c}}$

Hence, $\frac{1}{\sqrt{\lambda }}$ versus $Z$ is also a straight line.