Question

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

Answer 1

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 $m=\frac{m_0{c}^2}{\sqrt{1-v^2/c^2}}$, then

(a) Kinetic energy of electron = 1 eV = $1.6\times {10}^{-19}\mathrm{J}$

From eq. (1), we get

$$\begin{gathered} 1.6\times {10}^{-19}=\frac{m_0{c}^2}{\sqrt{1-v^2/c^2}}-m_0{c}^2 \\ \Rightarrow \frac{1.6\times {10}^{-19}}{m_0{c}^2}=\left(\frac{1}{\sqrt{1-v^2/c^2}}-1\right) \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{1}{\sqrt{1-v/c^2}}-1=\frac{1.6\times {10}^{-19}}{9.1\times {10}^{-31}\times 9\times {10}^{16}} \\ \Rightarrow \frac{1}{\sqrt{1-v/c^2}}-1=0.019536\times {10}^{-4} \\ \Rightarrow \frac{1}{\sqrt{1-v/c^2}}=1+0.019536\times {10}^{-4} \\ \Rightarrow \frac{1}{\sqrt{1-v/c^2}}=\frac{1}{1.0000019536} \\ \Rightarrow 1-v^2/c^2=0.99999613 \\ \Rightarrow v^2/c^2=0.00000387 \\ \Rightarrow v/c=0.001967231=3\times 0.001967231\times {10}^8 \\ =5.92\times {10}^5\mathrm{m}/\mathrm{s} \end{gathered}$$

(b) Kinetic energy of electron = 10 keV$=1.6\times {10}^{-19}\times 10\times {10}^3\mathrm{J}$

$$\begin{gathered} m_0{c}^2\left(\frac{1}{\sqrt{1-v^2/c^2}}-1\right)=1.6\times {10}^{-15} \\ \Rightarrow 9.1\times {10}^{-31}\times 9\times {10}^{16}\left(\frac{1}{\sqrt{1-v^2/c^2}}-1\right)=1.6\times {10}^{-15} \\ \Rightarrow \frac{1}{\sqrt{1-v^2/c^2}}-1=\frac{1.6\times {10}^{-15}}{9.1\times 9\times {10}^{-15}} \\ \Rightarrow \frac{1}{\sqrt{1-v^2/c^2}}-1=\frac{1.6}{9.1\times 9} \\ \Rightarrow \frac{1}{\sqrt{1-v^2/c^2}}=0.980838 \\ \Rightarrow 1-v^2/c^2=0.962043182 \\ \Rightarrow v^2/c^2=1-0.962043182 \\ \Rightarrow v^2=0.341611359\times {10}^{18} \\ \Rightarrow v=0.584475285\times {10}^8 \\ \Rightarrow v=5.85\times {10}^7\mathrm{m}/\mathrm{s} \end{gathered}$$

(c) Kinetic energy of electron$=10\mathrm{MeV}={10}^7\times 1.6\times {10}^{-19}\mathrm{J}$
$$\begin{gathered} \Rightarrow \frac{m_0{c}^2}{2\sqrt{1-v^2-c^2}}-m_0{c}^2=1.6\times {10}^{-12} \\ \Rightarrow m_0{c}^2\left(\frac{1}{\sqrt{1-v^2/c^2}}-1\right)=1.6\times {10}^{-12} \\ \Rightarrow \frac{1}{\sqrt{1-v^2\mathit{/}{c}^{\mathit{2}}}}-1=\frac{1.6\times {10}^{-12}}{9.1\times 9\times {10}^{-31}\times {10}^{16}} \\ \Rightarrow \frac{1}{\sqrt{1-v^2/c^2}}-1=1.019536\times {10}^3 \\ \Rightarrow \frac{1}{\sqrt{1-v^2/c^2}}-1=1019.536+1 \\ \Rightarrow \sqrt{1-v^2/c^2}=0.000979877 \\ \Rightarrow 1-v^2/c^2=0.99\times {10}^{-6} \\ \Rightarrow v=2.999999039\times {10}^8\mathrm{m}/\mathrm{s} \end{gathered}$$