Question

The wavelength of characteristic X - Ray $K_{\alpha }$ Line emitted from zinc (Z=30) is $\lambda$, find the wavelength of $K_{\alpha }$ Line emitted from Molybendum (Z=42).

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

According to Moseley’s law for $K_{\alpha }$ characteristic x rays, $\sqrt{v}=a(Z-b)$ where b=1.
Now, we know that x-ray being an electromagnetic wave; v = $\frac{c}{\lambda }$.
So replacing in the equation we get the relationship between the wavelength and the atomic number;
$\sqrt{\frac{c}{\lambda }}$ = $a(Z-1)$
or,
$\lambda$ = $\frac{c}{a^2(Z-1{)}^2}$
Now, we can have, for Zinc:
${\lambda }_{Zn}$ = $\frac{c}{a^2(20-1{)}^2}=\frac{c}{a^{2}341}$ ...(1) And, For Molybdenum:
${\lambda }_{Mo}$ = $\frac{c}{a^2(42-1{)}^2}=\frac{c}{a^{2}1651}$ ...(2) Dividing (2) by (1)
$\frac{{\lambda }_{Mo}}{{\lambda }_{Zn}}$ = $\frac{341}{1651}=0.5$ Hence,
${\lambda }_{Mo}$ = $\frac{{\lambda }_{Zn}}{2}$