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 (e/m for electron is ${1.810}^{11}$ C/kg)

• A. 1
• B. 0.1
• C. 0.2
• D. 0.3

Answer 1

The correct option is B 0.1

For incident electron,

$\frac{1}{2}{mV}^2=e^V$

$P^2=2meV$

$\therefore$ de - Bnogile wavelength ${\lambda }_1=\frac{h}{P}$

${\lambda }_1=\frac{h}{\sqrt{2meV}}$

Shortest $X-$rays wavelength, ${\lambda }_2=\frac{h_c}{e^V}$

$\therefore$ $\frac{{\lambda }_1}{{\lambda }_2}=\frac{1}{c}\sqrt{\left(\frac{V}{2}\right)\left(\frac{e}{m}\right)}$

$=\frac{\sqrt{\frac{{10}^4}{2}}\times 1.8\times {10}^{11}}{3\times {10}^8}=0.1$ is the correct answer.