Question

For modern X-ray tube, the wavelength of emitted X-rays and applied potential V of the tube are related as

• A. $\lambda =\frac{12400}{V}{A}^0$
• B. $\lambda =\frac{12400}{\sqrt{V}}{A}^0$
• C. $\lambda =\frac{1240}{\sqrt{V}}{A}^0$
• D. $\lambda =12400A^0$

Answer 1

The correct option is A $\lambda =\frac{12400}{V}{A}^0$

$\lambda =\frac{hc}{eV}$
$=\frac{20\times {10}^{-26}}{16\times {10}^{-19}\times V}$
$=\frac{12400}{V}{A}^{\circ }$