Question

The potential difference applied to an X-ray tube is V. The ratio of the de Broglie wavelength of electron to the minimum wavelength of X-ray is directly proportional to

• A. V
• B. $\sqrt{V}$
• C. $V^{3/2}$
• D. $V^{7/2}$

Answer 1

The correct option is B $\sqrt{V}$

The de Broglie wavelength of electron is ${\lambda }_e=\frac{h}{\sqrt{2meV}}$

The minimum wavelength of X-rays is ${\lambda }_X\left(min\right)=\frac{hc}{eV}$

Thus, $\frac{{\lambda }_e}{{\lambda }_X\left(min\right)}=\frac{h}{\sqrt{2meV}}\left(V/hc\right)\propto \sqrt{V}$