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Found in: Page 68

### Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087

# A proton is accelerated from rest through a potential difference of ${\mathbf{500}}{\mathbf{}}{\mathbit{M}}{\mathbit{V}}{\mathbf{.}}{\mathbf{}}$ (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 ${\mathbf{2000}}{\mathbf{}}{\mathbit{M}}{\mathbit{V}}$ potential difference? (c) What speed actually results?

1. At $500MV$, speed of a proton is $0.7567c$.
2. At $2000MV$, classical speed is $2c$
3. The relativistic speed is $0.9476c$.
See the step by step solution

## Step 1: Determine relation between kinetic energy and potential difference

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" $KE=∆U\phantom{\rule{0ex}{0ex}}KE=e∆V...................\left(1\right)$

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, m0 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)

## Step 2: (a) Speed when potential difference is500MV.

Consider the known data as below.

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

The speed of light, $c=3×{10}^{8}m}{s}$

The charge, $e=1.6×{10}^{-19}C$

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

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

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

Then using formula of Lorentz factor

localid="1659087345441"

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

## Step 3: (b) Classical speed when potential difference is quadrupled that is V=2000MV.

Here, we will equate classical kinetic energy to potential energy

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

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.

## Step 4: (c) Determine the relativistic speed to get the actual result:Rewrite equation (3) as below.

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

Here,

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

The speed of light, $c=3×{10}^{8}m}{s}$

The charge, $e=1.6×{10}^{-19}C$

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

Substitute all known numerical values in the above equation.

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

Then using formula of Lorentz factor as given by,

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