Question

If an X-ray tube operates at the voltage of 10 kV, find the ratio of the de broglie wavelength of the incident electrons to the shortest wavelength of X-rays produced. The specific charge of electron is $1.8\times {10}^{11}C\mathrm{kg.}$

• A. 1
• B. 0.1
• C. 1.8
• D. 1.2

Answer 1

The correct option is B 0.1

Kinetic energy gained by a charge q after being accelerated through a potential difference V volt, is given by
$qV=\frac{1}{2}m{v}^{2}mv=\sqrt{2mqV}$
As ${\lambda }_b=\frac{h}{mv}=\frac{h}{\sqrt{2qmv}}$
For cut-off wavelength of X-rays, we have

${\lambda }_m$=hc/qV
Now, $\frac{{\lambda }_b}{{\lambda }_m}$$=\frac{\sqrt{\frac{qV}{2m}}}{c}$
As $\frac{q}{m}=1.8\times {10}^{11}C{\mathrm{kg}}^{-1}\mathrm{for}\mathrm{electron}$
we have $\frac{{\lambda }_b}{{\mathrm{\lambda m}}^{}}=\frac{\sqrt{1.8\times {10}^{11}\times 10\times {10}^3/2}}{3\times {10}^8}=0.1$