# Find the Speed of an Electron with Kinetic Energy (A) 1 Ev, (B) 10 Kev and (C) 10 Mev. - Physics

Sum

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

#### Solution

If m0 is the rest mass of an electron and c is the speed of light, then kinetic energy of the electron = mc2 − m0c2       ...(1)
If $m = \frac{m_0 c^2}{\sqrt{1 - v^2 / c^2}}$

(a) Kinetic energy of electron = 1 eV = $1 . 6 \times {10}^{- 19} J$

From eq. (1), we get

$1 . 6 \times {10}^{- 19} = \frac{m_0 c^2}{\sqrt{1 - v^2 / c^2}} - m_0 c^2$

$\Rightarrow \frac{1 . 6 \times {10}^{- 19}}{m_0 c^2} = \left( \frac{1}{\sqrt{1 - v^2 / c^2}} - 1 \right)$

$\Rightarrow \frac{1}{\sqrt{1 - v/ c^2}} - 1 = \frac{1 . 6 \times {10}^{- 19}}{9 . 1 \times {10}^{- 31} \times 9 \times {10}^{16}}$
$\Rightarrow \frac{1}{\sqrt{1 - v/ c^2}} - 1 = 0 . 019536 \times {10}^{- 4}$
$\Rightarrow \frac{1}{\sqrt{1 - v/ c^2}} = 1 + 0 . 019536 \times {10}^{- 4}$
$\Rightarrow \frac{1}{\sqrt{1 - v/ c^2}} = \frac{1}{1 . 0000019536}$
$\Rightarrow 1 - v^2 / c^2 = 0 . 99999613$
$\Rightarrow v^2 / c^2 = 0 . 00000387$
$\Rightarrow v/c = 0 . 001967231 = 3 \times 0 . 001967231 \times {10}^8$
$= 5 . 92 \times {10}^5 m/s$

(b) Kinetic energy of electron = 10 keV

$= 1 . 6 \times {10}^{- 19} \times 10 \times {10}^3 J$

$m_0 c^2 \left( \frac{1}{\sqrt{1 - v^2 / c^2}} - 1 \right) = 1 . 6 \times {10}^{- 15}$
$\Rightarrow 9 . 1 \times {10}^{- 31} \times 9 \times {10}^{16} \left( \frac{1}{\sqrt{1 - v^2 / c^2}} - 1 \right) = 1 . 6 \times {10}^{- 15}$
$\Rightarrow \frac{1}{\sqrt{1 - v^2 / c^2}} - 1 = \frac{1 . 6 \times {10}^{- 15}}{9 . 1 \times 9 \times {10}^{- 15}}$
$\Rightarrow \frac{1}{\sqrt{1 - v^2 / c^2}} - 1 = \frac{1 . 6}{9 . 1 \times 9}$
$\Rightarrow \frac{1}{\sqrt{1 - v^2 / c^2}} = 0 . 980838$
$\Rightarrow 1 - v^2 / c^2 = 0 . 962043182$
$\Rightarrow v^2 / c^2 = 1 - 0 . 962043182$
$\Rightarrow v^2 = 0 . 341611359 \times {10}^{18}$
$\Rightarrow v = 0 . 584475285 \times {10}^8$
$\Rightarrow v = 5 . 85 \times {10}^7 m/s$

(c) Kinetic energy of electron

$= 10 MeV = {10}^7 \times 1 . 6 \times {10}^{- 19} J$

$\Rightarrow \frac{m_0 c^2}{2\sqrt{1 - v^2 - c^2}} - m_0 c^2 = 1 . 6 \times {10}^{- 12}$
$\Rightarrow m_0 c^2 \left( \frac{1}{\sqrt{1 - v^2 / c^2}} - 1 \right) = 1 . 6 \times {10}^{- 12}$
$\Rightarrow \frac{1}{\sqrt{1 - v^2 / c^2}} - 1 = \frac{1 . 6 \times {10}^{- 12}}{9 . 1 \times 9 \times {10}^{- 31} \times {10}^{16}}$
$\Rightarrow \frac{1}{\sqrt{1 - v^2 / c^2}} - 1 = 1 . 019536 \times {10}^3$
$\Rightarrow \frac{1}{\sqrt{1 - v^2 / c^2}} - 1 = 1019 . 536 + 1$
$\Rightarrow \sqrt{1 - v^2 / c^2} = 0 . 000979877$
$\Rightarrow 1 - v^2 / c^2 = 0 . 99 \times {10}^{- 6}$
$\Rightarrow v = 2 . 999999039 \times {10}^8 m/s$

Concept: Energy and Momentum
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 25 The Special Theory of Relativity
Q 26 | Page 458