Carol Valentine

2022-01-05

A proton with an initial speed of 800,000 m/s is brought to rest by an electric field. a. Did the proton move into a region of higher potential or lower potential? b. What was the potential difference that stopped the proton? c. What was the initial kinetic energy of the proton, in electron volts?

Lakisha Archer

Beginner2022-01-06Added 39 answers

Given th initial velocity of the proton $v}_{i}=800000\frac{m}{s$

a)The potencial is given by the formula:

$V=\frac{q}{U}$

The speed of the proton at the end is:

${v}_{f}=0$

As the proton has a positive charge and slows down while travelling, its potential is lower the beggining. Hence, the proton moves towards the region of higher potential.

b)We can get the potential difference from the law of energy conversation:

$K}_{i}+{U}_{i}={K}_{f}+{U}_{f$

$\frac{m{v}_{i}^{2}}{2}+q{V}_{i}=\frac{m{v}_{f}^{2}}{2}+q{V}_{f}$

$q{V}_{i}-q{V}_{f}=\frac{m{v}_{f}^{2}}{2}-\frac{m{v}_{i}^{2}}{2}$

$-q({V}_{f}-{V}_{i})=\frac{m{v}_{f}^{2}}{2}-\frac{m{v}_{i}^{2}}{2}$

$V=-\frac{\frac{m{v}_{f}^{2}}{2}-\frac{m{v}_{i}^{2}}{2}}{q}$

A proton has the following charge value

$q={e}^{+}=1.602\times {10}^{-19}C$

The speed of proton in the end is:

${v}_{f}=0$

The mass of the proton:

${m}_{p}=1673\times {10}^{-27}kg$

And we have

$V=-\frac{0-\frac{m{v}_{i}^{2}}{2}}{{e}^{-}}=\frac{1.673\times {10}^{-27}\times {\left(800000\right)}^{2}}{2\times 1.602\times {10}^{-19}}$

$V=3341.82V$

c)The initial kinetic energy of the proton is given by the equation

$K}_{i}=\frac{{m}_{p}{v}_{i}^{2}}{2$

${K}_{i}=\frac{1.673\times {10}^{-27}\times {\left(800000\right)}^{2}}{2}=5.3536\times {10}^{-16}J$

To conver kinetic energy into electron volts we can use:

$1eV=1.602\times {10}^{-19}J$

$1J=\frac{1}{1.602\times {10}^{-19}}eV$

$K}_{i}=5.3536\times {10}^{-16}\times \frac{1}{1.602\times {10}^{-19}$

${K}_{i}=3341.82eV$

a)The potencial is given by the formula:

The speed of the proton at the end is:

As the proton has a positive charge and slows down while travelling, its potential is lower the beggining. Hence, the proton moves towards the region of higher potential.

b)We can get the potential difference from the law of energy conversation:

A proton has the following charge value

The speed of proton in the end is:

The mass of the proton:

And we have

c)The initial kinetic energy of the proton is given by the equation

To conver kinetic energy into electron volts we can use:

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