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

Hint: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.