Question

In a Coolidge tube, high speed electrons emitted from the cathode are accelerated through a potential difference of $20kV$. Find the minimum wavelength of the X-rays produced.
(Take Planck's constant, $h=6.6\times {10}^{-34}Js$)

• A. 0.36782 A
• B. 0.61875 A
• C. 0.45867 A
• D. 0.72854 A

Answer 1

The correct option is B 0.61875 A

When electrons are accelerated through a potential difference, the work done on them is equal to the change in their kinetic energy. When cathode rays are stopped, they lose kinetic energy which is then converted into X-rays.
Thus, work done on cathode rays (W) $=V\times q=\frac{1}{2}m{v}^2$ $\to$ (1)
The energy and frequency of X-rays (any radiation) is related as $E=hv$. Where ${}^{\prime }{h}^{\prime }$ is Planck's constant.
But $v=\frac{c}{\lambda }$; $\therefore E=hv=\frac{hc}{\lambda }$ $\to$ (2)
Equating (1) and (2) and solving for $\lambda$.
$\frac{hc}{\lambda }=V\times q$; $\lambda =\frac{hc}{Vq}$. Substituting $h=6.6\times {10}^{-34}Js$, we have $c=3\times {10}^{8}m{s}^{-1}$.
$V=20\times {10}^{3}V,q=1.6\times {10}^{-19}C$
$\lambda =\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{20\times {10}^3\times 1.6\times {10}^{-19}}=0.61875A$