Question

In an X-ray tube the electrons are expected to strike the target with a velocity that is 10% of the velocity of light. The applied voltage should be:
(A) $517.6{\text{ }}V$
(B) $1052{\text{ }}V$
(C) $2.599{\text{ }}kV$
(D) $5.860{\text{ }}kV$

Answer 1

Kinetic energy of an electron of charge $e$ and potential difference $V$ is given as$K.E = eV$.Kinetic energy of an electron of mass $m$ and velocity $v$ is given as $K.E = \dfrac{1}{2}m{v^2}$

Complete step by step answer:
In an X-ray tube the electrons are generated in a cathode where the cathode is the negative terminal of an x-ray tube. When current flows electrons are emitted from the surface of the cathode by a process called thermionic emission. Then the electrons are attracted towards the positively charged anode and hit the target with a maximum energy determined by the tube potential (voltage).
The mass of the electron is given by ${m_e} = 9.1 \times {10^{ - 31}}kg$. The charge of electron is given by $e = 1.6 \times {10^{ - 19}}C$
Velocity of light is a constant quantity and given by $c = 3 \times {10^8}\dfrac{m}{s}$
The velocity of electron that strike the target is 10% of velocity of light i.e
$v = c \times 10\%$
$v = \dfrac{{10}}{{100}} \times 3 \times {10^8}\dfrac{m}{s} = 3 \times {10^7}\dfrac{m}{s}$
The relation between the applied voltage and the kinetic energy of the electron is given by $eV = \dfrac{1}{2}{m_e}{v^2}$, Where $V$ is the applied voltage.
$\Rightarrow V = \dfrac{{{m_e}{v^2}}}{{2e}} = \dfrac{{{m_e} = 9.1 \times {{10}^{ - 31}}kg \times {{(3 \times {{10}^7}m/s)}^2}}}{{2 \times 1.6 \times {{10}^{ - 19}}C}} = 2.599kV$
Hence the correct option is (C), the applied voltage is $2.599kV.$

Note: The most commonly used target material in the anode is tungsten because it has a high atomic number $\left( {Z = 74} \right)$ which increases the intensity of the x-rays.
Tungsten has a high melting point of $3370{\text{ }}^\circ C$ with a correspondingly low rate of evaporation.