Calculate the energy of one X-ray photon.

$$E_{\text{photon}} = hf$$ $$E_{\text{photon}} = (6.63 \times 10^{-34} \text{ Js}) \times (8 \times 10^{18} \text{ Hz})$$ $$E_{\text{photon}} = 5.304 \times 10^{-15} \text{ J}$$

$$E_{\text{photon}} = 0.80 \times KE_{\text{electron}}$$ $$KE_{\text{electron}} = \frac{E_{\text{photon}}}{0.80}$$ $$KE_{\text{electron}} = \frac{5.304 \times 10^{-15} \text{ J}}{0.80}$$ $$KE_{\text{electron}} = 6.63 \times 10^{-15} \text{ J}$$

$$KE_{\text{electron}} = \frac{1}{2} m_e v^2$$ $$v = \sqrt{\frac{2 \times KE_{\text{electron}}}{m_e}}$$ $$v = \sqrt{\frac{2 \times (6.63 \times 10^{-15} \text{ J})}{9.1 \times 10^{-31} \text{ kg}}}$$ $$v = \sqrt{1.45714 \times 10^{16}}$$ $$v \approx 1.21 \times 10^8 \text{ m/s}$$

The maximum velocity of the electrons is $\boxed{\text{1.21} \times 10^8 \text{ m/s}}$.

b) An X-ray tube operating at a P.D. of $50 \text{ kV}$ has a tube current $20 \text{ mA}$. Calculate:

(i) The electric input

(ii) The number of electrons hitting the target per second given that $e = 1.6 \times 10^{-19} \text{ C}$

(iii) The velocity of electrons when they hit the target.

Given:
Potential difference, $V = 50 \text{ kV} = 50 \times 10^3 \text{ V}$
Current, $I = 20 \text{ mA} = 20 \times 10^{-3} \text{ A}$
Charge of an electron, $e = 1.6 \times 10^{-19} \text{ C}$
Mass of electron, $m_e = 9.1 \times 10^{-31} \text{ kg}$

(i) The electric input