Question

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

• A. $0.1$
• B. $0.2$
• C. $0.3$
• D. $1.0$

Answer 1

Answer: (A)

$0.1$

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