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 V
• B. 1052 V
• C. 2.599 kV
• D. 5.860 kV

Answer 1

The correct option is C 2.599 kV

Given,

$m_e=9.1\times {10}^{-31}kg$

$e=1.6\times {10}^{-19}C$

$c=3\times {10}^{8}m/s$

The velocity of electron, $v=c.10$%

$v=\frac{10}{100}\times 3\times {10}^8=3\times {10}^{7}m/s$

The relation between the applied voltage an the kinetic energy of the electron,

$eV=\frac{1}{2}{m}_e{v}^2$

$V=\frac{m_e{v}^2}{2e}=\frac{9.1\times {10}^{-31}\times (3\times {10}^7{)}^2}{2\times 1.6\times {10}^{-19}}=2599V$

$V=2.599kV$

The correct option is C.