Question
Through what potential difference should an electron be accelerated to give it a speed of:
- 0.6c
- 0.9c
- 0.99c
✓
Answer
- $\text{eV}-\text{m}_0\text{C}^2=\frac{\text{m}_0\text{C}^2}{2\sqrt{1-\frac{\text{V}^2}{\text{C}^2}}}$
$=-\frac{9.1\times9\times10^{-31}\times10^{16}}{2\sqrt{1-\frac{0.36\text{C}^2}{\text{C}^2}}}$
$\Rightarrow\text{eV}-9.1\times9\times10^{-15}$
$=\frac{9.1\times9\times10^{-15}}{2\times0.08}$
$\Rightarrow\text{ev}-9.1\times9\times10^{-15}$
$=\frac{9.1\times9\times10^{-15}}{1.6}$
$\Rightarrow\text{eV}=\Big(\frac{9.1\times9}{1.6}+9.1\times9\Big)\times10^{-15}$
$=\text{eV}\Big(\frac{81.9}{1.6}+81.9\Big)\times10^{-15}$
$=\text{eV}=133.0875\times10^{-15}$
$\Rightarrow\text{V}=83.179\times10^{4}=831\text{KV}$
- $\text{eV}=\text{m}_0\text{C}^2=\frac{\text{m}_0\text{C}^2}{2\sqrt{1-\frac{\text{V}^2}{\text{C}^2}}}$
$=\frac{9.1\times9\times10^{-15}}{2\sqrt{1-\frac{0.81\text{C}^2}{\text{C}^2}}}$
$\Rightarrow\text{eV}-81.9\times10^{-15}=\frac{9.1\times9\times10^{-15}}{2\times0.435}$
$\Rightarrow\text{eV}=12.237\times10^{-15}$
$\Rightarrow\text{V}=\frac{12.237\times10^{-15}}{1.6\times10^{-19}}=76.48\text{kV}$
$\text{V}=0.99\text{C}=\text{ev}-\text{m}_0\text{C}^2=\frac{\text{m}_0\text{C}^2}{2\sqrt{1-\frac{\text{V}^2}{\text{C}^2}}}$
$\Rightarrow\text{eV}=\frac{\text{m}_0\text{C}^2}{2\sqrt{1-\frac{\text{V}^2}{\text{C}^2}}}+\text{m}_0\text{C}^2$
$=\frac{9.1\times10^{31}\times9\times10^{16}}{2\sqrt{1-(0.99)^2}} +9.1\times10^{-31}\times9\times10^{16}$
$\Rightarrow\text{eV}=372.18\times10^{-15}$
$\Rightarrow\text{V}=\frac{372.18\times20^{-15}}{1.6\times10^{-19}}=272.6\times10^{4}$
$\Rightarrow\text{V}=2.726\times10^{6}=2.7\text{MeV}$
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