Question

Find the speed of an electron with kinetic energy (a) 1 eV, (b) 10 keV and (c) 10 MeV.

Answer 1

Answer:

can you please give value of kinetic energy

Answer 2

Explanation:

If c is the light’s speed,$m_0$ is the electron’s rest mass, then the electron’s kinetic energy = $m c^{2}-m_{0} c^{2}$       …(1)

If m = $m=m 0 c 21-v 2 / c 2$, then

(a) Electron’s kinetic energy = $1 \mathrm{eV}=1.6 \times 10-19 \mathrm{J}$

From equation (1), we obtain

$1.6 \times 10-19=m 0 c 21-v 2 / c 2-m 0 c 2$

$\Rightarrow \mathrm{v} / \mathrm{c}=3 \times 0.001967231 \times 108=5.92 \times 105 \mathrm{m} / \mathrm{s}$

(b) Electron’s kinetic energy = 10 keV

$=1.6 \times 10 \times 10-19 \times 103 \mathrm{J} \mathrm{moc} 211-\mathrm{v} 2 / \mathrm{c} 2-1=1.6 \times 10^{-15}$

$\Rightarrow 11-\mathrm{v} 2 / \mathrm{c} 2-1=1.69 .1 \times 9$

$\Rightarrow v=5.85 \times 107 \mathrm{m} / \mathrm{s}$

(c) Electron’s kinetic energy  

$=10 \mathrm{MeV}=107 \times 10-191 \times 1.6$

$\Rightarrow \mathrm{moc} 221-\mathrm{v} 2-\mathrm{c} 2-\mathrm{moc} 2=1.6 \times 10-12$

$\Rightarrow 1-\mathrm{v} 2 / \mathrm{c} 2=0.99 \times 10-6$

$\Rightarrow v=2.999999039 \times 108 \mathrm{m} / \mathrm{s}$