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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.

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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
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