Question

The wave length of $K_{\alpha }$, x- rays produced by an x-ray tube is $\lambda$. The atomic number of the anode material of the tube is: (Constants have their usual meanings)

• A. $\frac{1}{a}\sqrt{\frac{c}{\lambda }}+1$
• B. $\frac{1}{a}\sqrt{\frac{c}{\lambda }}-1$
• C. $a\sqrt{\frac{c}{\lambda }}+1$
• D. $a\sqrt{\frac{c}{\lambda }}-1$

Answer 1

The correct option is A $\frac{1}{a}\sqrt{\frac{c}{\lambda }}+1$

The famous empirical formula of Moseley which led foundation to the most modern periodic table.
It says that the square root of frequency of $K_{\alpha }$ characteristic x-rays depends on atomic number of the anode atom linearly.
Mathematically represented as;
$\sqrt{v}$ = a (Z-b)
Where a and b are constants and b=1.
Now,
X-ray being an electromagnetic photon;
$v=\frac{c}{\lambda }$
Which leads us to the following equation
$\sqrt{\frac{c}{\lambda }}=a(Z-1)$ ( Here $\lambda$ is the wavelength of characteristic $K_{\alpha }$ x-ray mentioned in the problem)
We need Z, hence by rearranging the equation we get;
$Z=\frac{1}{a}\sqrt{\frac{c}{\lambda }}+1$