Question

This one will need you to work the calculator a little bit. Given that an x-ray tube operates at 20Kv, what is the maximum speed of the electrons striking the anode? Take $1eV=1.6\times {10}^{19}$ J and $m_e=9\times {10}^{-31}$kg

• A. $4.2\times {10}^{8}m/s$
• B. $9.4\times {10}^{8}m/s$
• C. $8.4\times {10}^{7}m/s$
• D. $5.1\times {10}^{7}m/s$

Answer 1

The correct option is C $8.4\times {10}^{7}m/s$

An electron accelerated through a potential difference of 20kVwill gain 20keVor 20,00eVof energy. In Joules,
$20,000eV=20,000\times 1.6\times {10}^{-19}J=3.2\times {10}^{-15}J.$
This will be converted into kinetic energy of the electrons -
$\frac{m_e{v}^2}{2}=3.2\times {10}^{-15}J\Rightarrow V=\sqrt{\frac{2\times (3.2\times {10}^{-15}J)}{9\times {10}^{-31}kg}}=8.4\times {10}^{7}m/s.$