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Q87E

Expert-verifiedFound in: Page 68

Book edition
2nd Edition

Author(s)
Randy Harris

Pages
633 pages

ISBN
9780805303087

** A proton is accelerated from rest through a potential difference of ${\mathbf{500}}{\mathbf{}}{\mathit{M}}{\mathit{V}}{\mathbf{.}}{\mathbf{}}$ **

**(a) What is its speed? **

**(b) Classical mechanics indicates that quadrupling the potential difference would double the speed. Were a classical analysis valid, what speed would result from a ${\mathbf{2000}}{\mathbf{}}{\mathit{M}}{\mathit{V}}$ potential difference? **

**(c) What speed actually results?**

- At $500MV$, speed of a proton is $0.7567c$.
- At $2000MV$, classical speed is $2c$
- The relativistic speed is $0.9476c$.

To find the speed of a proton accelerated under a potential difference of $500MV$, we must equate relativistic kinetic energy to change in potential energy.

**Relativistic kinetic energy is equal to change in potential energy as the charged particle is accelerated under a potential difference**

localid="1659086760380" $KE=\u2206U\phantom{\rule{0ex}{0ex}}KE=e\u2206V...................\left(1\right)$

Here, KE is the kinetic energy, $\Delta U$ is the change in potential energy, e is the charge, and $\Delta V$ is the potential difference.

And as the electron is traveling at a very high speed considering relativistic Kinetic energy as below.

$\begin{array}{rcl}KE& =& Totalenergy-{m}_{o}{c}^{2}\\ & =& m{c}^{2}-{m}_{o}{c}^{2}\\ & =& \gamma {m}_{o}{c}^{2}-{m}_{o}{c}^{2}\\ & & \end{array}$

$KE=\left(\gamma -1\right){m}_{o}{c}^{2}$ ….. (2)

Here, m_{0} is the rest mass which is equal to the mass of the proton. Therefore, by comparing equation (1) and (2), you get

$\left(\gamma -1\right){m}_{p}{c}^{2}=e\Delta V$ ….. (3)

Consider the known data as below.

The mass of proton, ${m}_{p}=1.67\times {10}^{-27}kg$

The speed of light, $c=3\times {10}^{8}\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

The charge, $e=1.6\times {10}^{-19}C$

The potential difference, $\Delta V=500\text{}MV=5\times {10}^{8}V$

Substitute all known numerical values into equation (3), and you have

$(\gamma -1)(1.67\times {10}^{-27}kg){(3\times {10}^{8}\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.)}^{2}=(1.6\times {10}^{-19}C)(5\times {10}^{8}V)\phantom{\rule{0ex}{0ex}}(\gamma -1)=0.532\phantom{\rule{0ex}{0ex}}\gamma =1.532$

Then using formula of Lorentz factor

localid="1659087345441" $\gamma =\frac{1}{\sqrt{1-\raisebox{1ex}{${v}^{2}$}\!\left/ \!\raisebox{-1ex}{${c}^{2}$}\right.}}\phantom{\rule{0ex}{0ex}}1.532=\frac{1}{\sqrt{1-\raisebox{1ex}{${v}^{2}$}\!\left/ \!\raisebox{-1ex}{${c}^{2}$}\right.}}\phantom{\rule{0ex}{0ex}}1-\raisebox{1ex}{${v}^{2}$}\!\left/ \!\raisebox{-1ex}{${c}^{2}$}\right.=\frac{1}{2.347}\phantom{\rule{0ex}{0ex}}\raisebox{1ex}{${v}^{2}$}\!\left/ \!\raisebox{-1ex}{${c}^{2}$}\right.=1-0.426\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{v}^{2}=0.574{c}^{2}\phantom{\rule{0ex}{0ex}}v=0.758c\phantom{\rule{0ex}{0ex}}$

Hence, at 500MV , the peed of a proton is 0.758c.

Here, we will equate classical kinetic energy to potential energy

$\frac{1}{2}m{v}^{2}=eV\phantom{\rule{0ex}{0ex}}{v}^{2}=\frac{2eV}{m}\phantom{\rule{0ex}{0ex}}{v}^{2}=\frac{2\left(1.6\times {10}^{-19}C\right)\left(2\times {10}^{9}\right)}{1.67\times {10}^{-27}}\phantom{\rule{0ex}{0ex}}v\approx 2c$

Hence, at ,2000MV the classical speed is 2c .

According to the Einstein’s second postulate that, the speed of light is constant irrespective of the reference frame.

Therefore, classical result is invalid.

$\left(\gamma -1\right){m}_{p}{c}^{2}=e\Delta V$

Here,

The mass of proton, ${m}_{p}=1.67\times {10}^{-27}kg$

The speed of light, $c=3\times {10}^{8}\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

The charge, $e=1.6\times {10}^{-19}C$

The potential difference, $\Delta V=2000\text{}MV=2\times {10}^{9}V$

** **

Substitute all known numerical values in the above equation.

$\left(\gamma -1\right)\left(1.67\times {10}^{-27}kg\right){\left(3\times {10}^{8}\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)}^{2}=\left(1.6\times {10}^{-19}C\right)\left(2\times {10}^{9}V\right)\phantom{\rule{0ex}{0ex}}\left(\gamma -1\right)\left(15.03\times {10}^{-11}\right)=\left(3.2\times {10}^{-10}\right)\phantom{\rule{0ex}{0ex}}\left(\gamma -1\right)=2.129\phantom{\rule{0ex}{0ex}}\gamma =3.129$

Then using formula of Lorentz factor as given by,

$\gamma =\frac{1}{\sqrt{1-\raisebox{1ex}{${v}^{2}$}\!\left/ \!\raisebox{-1ex}{${c}^{2}$}\right.}}\phantom{\rule{0ex}{0ex}}3.129=\frac{1}{\sqrt{1-\raisebox{1ex}{${v}^{2}$}\!\left/ \!\raisebox{-1ex}{${c}^{2}$}\right.}}\phantom{\rule{0ex}{0ex}}1-\raisebox{1ex}{${v}^{2}$}\!\left/ \!\raisebox{-1ex}{${c}^{2}$}\right.=\frac{1}{9.791}\phantom{\rule{0ex}{0ex}}\raisebox{1ex}{${v}^{2}$}\!\left/ \!\raisebox{-1ex}{${c}^{2}$}\right.=1-0.102\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{v}^{2}=0.898{c}^{2}\phantom{\rule{0ex}{0ex}}v=0.947c\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Hence, at 2000MV , the Speed of a proton is 0.947c.

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