Question

$K_{\alpha }$ wavelength emitted by an atom of atomic number Z = 11 is $\lambda$. Find the atomic number for an atom that emits $K_{\alpha }$ radiation with wavelength $4\lambda$

• A. Z = 6
• B. Z = 4
• C. Z = 11
• D. Z = 44

Answer 1

The correct option is A Z = 6

$\sqrt{f_1}=\sqrt{\frac{v}{{\lambda }_1}}=\alpha (11-1)$ and $\sqrt{f_2}=\sqrt{\frac{v}{{\lambda }_2}}=\alpha (Z-1)$
By dividing, $\sqrt{\frac{{\lambda }_2}{{\lambda }_1}}=\frac{10}{Z=1}\Rightarrow \sqrt{\frac{4}{1}}=\frac{10}{Z=1}\Rightarrow Z=6$