Question

The anode (positive terminal) of an X-ray tube is at a potential of 120,000 V with respect to the cathode (negative terminal). How much work (in Joules) is done by the electric force when an electron is accelerated from the anode to the cathode? If the electron is initially at rest, find the speed of the electron when it arrives at the cathode. (e = 1.6 * 10^-19 C,     me =9.1 * 10^-31 kg)

Answer 1

q = 1.6 x 10^-19 C

V = potential difference = 120000 volt

rest mass m_o = 9.1 x 10^-31 kg

$m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$

rest mass energy of electron m_oc² = 511 keV

kinetic energy of electron when accelerated through 120000 volt = 120 keV

total energy E = kinetic energy + rest mass energy = 631 keV

$pc = \sqrt{E^2-m_0^2c^4}$ = 370.2 keV

$pc =mvc= \left(\frac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}} \right )\left(\frac{v}{c} \right )=Ev/c$

v/c = 370.2/631 = 0.587

v = 0.587 x 3 x 10⁸ = 1.76 x 10⁸ m/s