Question

An X-ray tube, operated at a potential difference of 40$kV$ , produces heat at the rate of 720$W$ . Assuming 0.5$\%$ of the energy of incident electrons is converted into X-rays, calculate
(i) the number of electron per second striking the target (ii) the velocity of the incident electrons

Answer 1

$\left(a\right)$ Heat produced per second at the target

$P=0.995VI(0.5\%ofenergyisconvertedintox-rays)$

$\Rightarrow I=\frac{P}{0.995V}=\frac{720}{0.995V}$

$=\frac{720}{0.995\times 40\times {10}^3}$

$=0.018ampere$

The no. of $e-$emitted per second :

$n=\frac{I}{e}$

$=\frac{0.018}{1.6\times {10}^{-19}}$

$=1.1\times {10}^{17}$ electron

$\left(b\right)$ Energy of photon $\frac{1}{2}m{v}^2=eV$

$\Rightarrow v=\frac{\sqrt{2eV}}{m}=1.2\times {10}^{8}m/s$