Question

Atomic number of anticathode material in an $X$-ray tube is $41$ . Wavelength of $K_{\alpha }X$-ray produced in the tube is

• A. $0.66\mathring{A}$
• B. $0.76\mathring{A}$
• C. $0.82\mathring{A}$
• D. $0.88\mathring{A}$

Answer 1

Answer: (B)

$\sqrt{f}=K_{\alpha }(Z-1)$
$\sqrt{f}=(40)$
$\sqrt{f}=\sqrt{2.48\times 10^{15}}\times 40$
$f=2.48\times 10^{15}\times 1600$
$\frac{c}{\lambda }=2.48\times 10^{15}\times 1600$
$\lambda =0.76\times 10^{-10}$
or $\lambda =0.76\mathring{A}$