Question

If 10000 V is applied across an X-ray tube, what will be the ratio of de-Broglie wavelength of the incident electrons to the shortest wavelength of X-ray produced ( $\frac{e}{m}$ for electron is $1.8\times {10}^{11}ck{g}^{-1}$ )

• A. 1
• B. 0.1
• C. 0.2
• D. 0.3

Answer 1

The correct option is B 0.1

For the incident electron $\frac{1}{2}m{v}^2=eVor{p}^2=2meV$
$\therefore$ de-Broglie wavelength ${\lambda }_1=\frac{h}{p}=\frac{h}{\sqrt{2meV}}$
Shortest X-ray wavelength ${\lambda }_2=\frac{hc}{eV}$
$\therefore \frac{{\lambda }_1}{{\lambda }_2}=\frac{1}{c}\sqrt{\left(\frac{V}{2}\right)\left(\frac{e}{m}\right)}=\frac{\sqrt{\frac{{10}^4}{2}\times 1.8\times {10}^{11}}}{3\times {10}^8}=0.1$