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

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:
${K}_{i}+{U}_{i}={K}_{f}+{U}_{f}$
$\frac{m{v}_{i}^{2}}{2}+q{V}_{i}=\frac{m{v}_{f}^{2}}{2}+q{V}_{f}$
$q{V}_{i}-q{V}_{f}=\frac{m{v}_{f}^{2}}{2}-\frac{m{v}_{i}^{2}}{2}$
$-q\left({V}_{f}-{V}_{i}\right)=\frac{m{v}_{f}^{2}}{2}-\frac{m{v}_{i}^{2}}{2}$
$V=-\frac{\frac{m{v}_{f}^{2}}{2}-\frac{m{v}_{i}^{2}}{2}}{q}$
A proton has the following charge value
$q={e}^{+}=1.602×{10}^{-19}C$
The speed of proton in the end is:
${v}_{f}=0$
The mass of the proton:
${m}_{p}=1673×{10}^{-27}kg$
And we have
$V=-\frac{0-\frac{m{v}_{i}^{2}}{2}}{{e}^{-}}=\frac{1.673×{10}^{-27}×{\left(800000\right)}^{2}}{2×1.602×{10}^{-19}}$
$V=3341.82V$
c)The initial kinetic energy of the proton is given by the equation
${K}_{i}=\frac{{m}_{p}{v}_{i}^{2}}{2}$
${K}_{i}=\frac{1.673×{10}^{-27}×{\left(800000\right)}^{2}}{2}=5.3536×{10}^{-16}J$
To conver kinetic energy into electron volts we can use:
$1eV=1.602×{10}^{-19}J$
$1J=\frac{1}{1.602×{10}^{-19}}eV$
${K}_{i}=5.3536×{10}^{-16}×\frac{1}{1.602×{10}^{-19}}$
${K}_{i}=3341.82eV$

Vasquez

a. To determine if the proton moved into a region of higher or lower potential, we need to consider the change in electric potential energy of the proton. If the potential energy decreased, then the proton moved into a region of lower potential. If the potential energy increased, then the proton moved into a region of higher potential.
Since the proton is brought to rest by the electric field, all of its initial kinetic energy is converted into electric potential energy. Therefore, the potential energy of the proton increased, and it moved into a region of higher potential.
b. The potential difference that stopped the proton is equal to the change in electric potential energy of the proton. We can use the formula:
$\Delta U=q\Delta V$
where $\Delta U$ is the change in electric potential energy of the proton, $q$ is the charge of the proton (which is the elementary charge $e$), and $\Delta V$ is the potential difference that stopped the proton.
Since the proton is brought to rest, its final kinetic energy is zero. Therefore, the change in electric potential energy is equal to its initial kinetic energy:
$\Delta U={K}_{i}=\frac{1}{2}m{v}_{i}^{2}$
where $m$ is the mass of the proton, ${v}_{i}$ is the initial speed of the proton, and ${K}_{i}$ is the initial kinetic energy of the proton.
Substituting the given values, we get:

Therefore, the potential difference that stopped the proton is:

c. To find the initial kinetic energy of the proton in electron volts (eV), we can use the conversion factor:

Therefore, the initial kinetic energy of the proton is:

Converting to eV, we get:

Therefore, the initial kinetic energy of the proton is 3.34 million electron volts.

user_27qwe

Step 1:
a. To determine whether the proton moved into a region of higher or lower potential, we need to analyze its initial speed. Since the proton was brought to rest, we can conclude that it moved into a region of higher potential.
Step 2:
b. The potential difference required to stop the proton can be calculated using the equation: $\Delta V=\frac{1}{q}\Delta KE$, where $\Delta V$ is the potential difference, $q$ is the charge of the proton, and $\Delta KE$ is the change in kinetic energy. The charge of a proton is $q=1.6×{10}^{-19}$ C. The proton was brought to rest, so its change in kinetic energy is equal to its initial kinetic energy. Thus, $\Delta KE=K{E}_{\text{initial}}$. Plugging in the values, we have:
$\Delta V=\frac{1}{1.6×{10}^{-19}\phantom{\rule{0.167em}{0ex}}\text{C}}×K{E}_{\text{initial}}$
Step 3:
c. To find the initial kinetic energy of the proton in electron volts (eV), we can use the relationship: $1\phantom{\rule{0.167em}{0ex}}\text{eV}=1.6×{10}^{-19}$ J. The initial kinetic energy of the proton can be calculated using the formula: $K{E}_{\text{initial}}=\frac{1}{2}m{v}_{\text{initial}}^{2}$, where $m$ is the mass of the proton and ${v}_{\text{initial}}$ is its initial speed. The mass of a proton is $m=1.67×{10}^{-27}$ kg. Substituting the values, we have:
$K{E}_{\text{initial}}=\frac{1}{2}×\left(1.67×{10}^{-27}\phantom{\rule{0.167em}{0ex}}\text{kg}\right)×\left(800,000\phantom{\rule{0.167em}{0ex}}\text{m/s}{\right)}^{2}$

karton

To solve this problem, let's consider the electric potential energy of a proton in an electric field. The electric potential energy ($U$) of a proton is given by the equation:
$U=q·V$ where $q$ is the charge of the proton and $V$ is the electric potential.
a. To determine whether the proton moved into a region of higher potential or lower potential, we need to compare the initial potential (${V}_{\text{initial}}$) with the final potential (${V}_{\text{final}}$) when the proton comes to rest.
If the proton comes to rest, it means that its initial kinetic energy is completely converted into potential energy. Therefore, we can say that ${U}_{\text{initial}}={U}_{\text{final}}$.
The initial kinetic energy of the proton can be calculated using the equation:
${K}_{\text{initial}}=\frac{1}{2}m{v}_{\text{initial}}^{2}$ where $m$ is the mass of the proton and ${v}_{\text{initial}}$ is its initial speed.
b. The potential difference ($\Delta V$) that stopped the proton can be found by subtracting the final potential from the initial potential:
$\Delta V={V}_{\text{final}}-{V}_{\text{initial}}$
c. To calculate the initial kinetic energy of the proton in electron volts (eV), we need to convert the mass of the proton from kilograms (kg) to electron volts (eV). The conversion factor is given by:
and the mass of a proton is approximately $1.67×{10}^{-27}$ kg.
Now let's calculate the solutions using the given values and equations:
a. The proton came to rest, so we can conclude that it moved into a region of .
b. We can use the equation $\Delta V={V}_{\text{final}}-{V}_{\text{initial}}$ to find the potential difference that stopped the proton.
Since the proton came to rest, the final potential is zero (${V}_{\text{final}}=0$). Therefore:
$\Delta V=0-{V}_{\text{initial}}=-{V}_{\text{initial}}$
The potential difference that stopped the proton is $\overline{)-{V}_{\text{initial}}}$.
c. First, let's convert the mass of the proton to electron volts (eV):

Now, we can calculate the initial kinetic energy in eV using the equation:
${K}_{\text{initial}}=\frac{1}{2}m{v}_{\text{initial}}^{2}$
Substituting the given values:

Finally, we convert the result to eV:

Simplifying this expression will give us the initial kinetic energy of the proton in eV.

star233

a. Did the proton move into a region of higher potential or lower potential?
Since the proton was brought to rest, it means its kinetic energy decreased. Therefore, the proton moved into a region of .
b. What was the potential difference that stopped the proton?
Using the equation $\Delta V=\frac{\Delta KE}{q}$, and considering the proton's final kinetic energy is zero, we have:
$\Delta V=-\frac{K{E}_{\text{initial}}}{q}$
Substituting the values, we get:
$\Delta V=-\frac{\frac{1}{2}m{v}^{2}}{q}$
To calculate the potential difference, we need to know the mass and charge of the proton. The mass of a proton, $m$, is approximately $1.67×{10}^{-27}$ kg, and its charge, $q$, is $1.6×{10}^{-19}$ C.
Substituting these values and the given initial speed, $v$, we can calculate the potential difference, $\Delta V$:
$\Delta V=-\frac{\frac{1}{2}×1.67×{10}^{-27}×\left(800,000{\right)}^{2}}{1.6×{10}^{-19}}$
Evaluating this expression gives the value of $\Delta V$ in volts.
c. What was the initial kinetic energy of the proton, in electron volts?
The initial kinetic energy of the proton can be calculated using the equation:
$KE=\frac{1}{2}m{v}^{2}$
Substituting the values for the mass, $m$, and the initial speed, $v$, we can calculate the initial kinetic energy, $KE$, in joules.
To convert this value to electron volts (eV), we use the conversion factor: 1 eV = $1.6×{10}^{-19}$ J. Therefore, we can convert the initial kinetic energy from joules to electron volts by dividing it by $1.6×{10}^{-19}$.
$K{E}_{\text{eV}}=\frac{K{E}_{\text{initial}}}{1.6×{10}^{-19}}$
Substituting the calculated value of $K{E}_{\text{initial}}$ into the equation gives the initial kinetic energy in electron volts.

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