David Lewis

2022-01-03

An electron with an initial speed of $6.00×{10}^{5}$ m/sm/s is brought to rest by an electric field.
a)Did the electron move into a region of higher potential or lower potential?
b)What was the potential difference that stopped the electron?
c)What was the initial kinetic energy of the electron, in electron volts?

Anzante2m

(a) Electron moves into a region of lower potential because it must moves in the opposite direction of force to get in rest.
(b) The stopping potential difference for the electrons is,
$\frac{m{v}^{2}}{2}=-q\mathrm{△}V$
$\mathrm{△}V=\frac{m{v}^{2}}{2e}$
Substitute the values,
Substitute the values,
$\mathrm{△}V=\frac{\left(9.11×{10}^{-31}kg\right){\left(6.00×{10}^{5}\frac{m}{s}\right)}^{2}}{2\left(1.602×{10}^{-19}C\right)}$
$=102.4×{10}^{-2}$ V
$\approx 1.02$ V
c) The initial kinetic energy of the electrons in electron volt is,
$\mathrm{△}K=\frac{m{v}^{2}}{2}$
$={\frac{9.11×{10}^{-31}kg}{6.00×{10}^{5}\frac{m}{s}}}^{2}\right\}\left\{2\right\}$
$=1.64×{10}^{-19}J\left(\frac{1eV}{1.602×{10}^{-19}J}\right)$
$=1.02eV$

Jazz Frenia

Step 1:
a) Did the electron move into a region of higher potential or lower potential?
An electron with an initial speed of $6.00×{10}^{5}$ m/s is brought to rest. When an electron is brought to rest, it means that all its initial kinetic energy has been converted into electric potential energy. Thus, the electron moved into a region of lower potential.
Step 2:
b) What was the potential difference that stopped the electron?
The potential difference ($\Delta V$) can be calculated using the equation:

where $q$ is the charge of the electron, which is $-1.6×{10}^{-19}$ C.
Since the electron is brought to rest, its final kinetic energy is zero. Therefore, we can write:
$\Delta V=\frac{0}{-1.6×{10}^{-19}\phantom{\rule{0.167em}{0ex}}\text{C}}$
$\Delta V=0\phantom{\rule{0.167em}{0ex}}\text{{V}\right\}$
Hence, the potential difference that stopped the electron is 0 V.
Step 3:
c) What was the initial kinetic energy of the electron, in electron volts?
The initial kinetic energy of the electron can be calculated using the equation:
$K=\frac{1}{2}m{v}^{2}$
where $m$ is the mass of the electron, which is $9.11×{10}^{-31}$ kg, and $v$ is the initial speed of the electron, which is $6.00×{10}^{5}$ m/s.
Substituting these values into the equation, we have:
$K=\frac{1}{2}×9.11×{10}^{-31}\phantom{\rule{0.167em}{0ex}}\text{{kg}\right\}×\left(6.00×{10}^{5}\phantom{\rule{0.167em}{0ex}}\text{{m/s}\right\}{\right)}^{2}$
Simplifying this expression will give us the initial kinetic energy in joules. However, we are asked to express it in electron volts (eV). To convert joules to electron volts, we can use the conversion factor:
$1\phantom{\rule{0.167em}{0ex}}\text{{eV}\right\}=1.6×{10}^{-19}\phantom{\rule{0.167em}{0ex}}\text{{J}\right\}$
Let's calculate the initial kinetic energy in joules and then convert it to electron volts:
$K=\frac{1}{2}×9.11×{10}^{-31}\phantom{\rule{0.167em}{0ex}}\text{{kg}\right\}×\left(6.00×{10}^{5}\phantom{\rule{0.167em}{0ex}}\text{{m/s}\right\}{\right)}^{2}$
$K=8.199×{10}^{-19}\phantom{\rule{0.167em}{0ex}}\text{{J}\right\}$
Converting to electron volts:
${K}_{\text{eV}}=\frac{8.199×{10}^{-19}\phantom{\rule{0.167em}{0ex}}\text{J}}{1.6×{10}^{-19}\phantom{\rule{0.167em}{0ex}}\text{J/eV}}$
${K}_{\text{eV}}=5.124\phantom{\rule{0.167em}{0ex}}\text{{eV}\right\}$
Therefore, the initial kinetic energy of the electron is $5.124$ eV.

Nick Camelot

a) To determine whether the electron moved into a region of higher or lower potential, we can consider its initial speed. If the electron was brought to rest, it must have lost kinetic energy, which means it moved into a region of lower potential.
b) The potential difference required to stop the electron can be calculated using the equation:
$\Delta V=\frac{\Delta K}{q}$
where $\Delta V$ is the potential difference, $\Delta K$ is the change in kinetic energy, and $q$ is the charge of the electron.
Since the electron was brought to rest, its final kinetic energy is zero. Thus, the change in kinetic energy ($\Delta K$) is equal to the negative of its initial kinetic energy (${K}_{i}$):
$\Delta K=-{K}_{i}$
Therefore, the potential difference can be expressed as:
$\Delta V=\frac{-{K}_{i}}{q}$
c) The initial kinetic energy of the electron can be calculated using the equation:
${K}_{i}=\frac{1}{2}m{v}^{2}$
where $m$ is the mass of the electron and $v$ is its initial velocity.
Now, let's calculate the values using the given information:
Given:
Initial speed of the electron, $v=6.00×{10}^{5}\phantom{\rule{0.167em}{0ex}}\text{m/s}$
Charge of the electron, $q=1.6×{10}^{-19}\phantom{\rule{0.167em}{0ex}}\text{C}$
Mass of the electron, $m=9.11×{10}^{-31}\phantom{\rule{0.167em}{0ex}}\text{kg}$
We can substitute these values into the equations:
${K}_{i}=\frac{1}{2}×\left(9.11×{10}^{-31}\phantom{\rule{0.167em}{0ex}}\text{kg}\right)×\left(6.00×{10}^{5}\phantom{\rule{0.167em}{0ex}}\text{m/s}{\right)}^{2}$
Now we can calculate ${K}_{i}$:
${K}_{i}=\frac{1}{2}×\left(9.11×{10}^{-31}\phantom{\rule{0.167em}{0ex}}\text{kg}\right)×\left(6.00×{10}^{5}\phantom{\rule{0.167em}{0ex}}\text{m/s}{\right)}^{2}$
Simplifying the calculation:
${K}_{i}=\frac{1}{2}×\left(9.11×{10}^{-31}\phantom{\rule{0.167em}{0ex}}\text{kg}\right)×\left(36.00×{10}^{10}\phantom{\rule{0.167em}{0ex}}{\text{m}}^{2}/{\text{s}}^{2}\right)$
${K}_{i}=0.5×9.11×36.00×{10}^{-21}\phantom{\rule{0.167em}{0ex}}\text{kg}·{\text{m}}^{2}/{\text{s}}^{2}$
${K}_{i}=164.34×{10}^{-21}\phantom{\rule{0.167em}{0ex}}\text{kg}·{\text{m}}^{2}/{\text{s}}^{2}$
Now, to express the initial kinetic energy in electron volts (eV), we can use the conversion factor:
$1\phantom{\rule{0.167em}{0ex}}\text{eV}=1.6×{10}^{-19}\phantom{\rule{0.167em}{0ex}}\text{J}$
Thus, we can convert the initial kinetic energy from joules to electron volts by dividing it by the conversion factor:
${K}_{i}\phantom{\rule{0.167em}{0ex}}\left(\text{eV}\right)=\frac{164.34×{10}^{-21}\phantom{\rule{0.167em}{0ex}}\text{kg}·{\text{m}}^{2}/{\text{s}}^{2}}{1.6×{10}^{-19}\phantom{\rule{0.167em}{0ex}}\text{J/eV}}$
Simplifying the calculation:
${K}_{i}\phantom{\rule{0.167em}{0ex}}\left(\text{eV}\right)=\frac{164.34}{1.6}×{10}^{-21-\left(-19\right)}\phantom{\rule{0.167em}{0ex}}\text{eV}$
${K}_{i}\phantom{\rule{0.167em}{0ex}}\left(\text{eV}\right)=102.71\phantom{\rule{0.167em}{0ex}}\text{eV}$
Therefore, the initial kinetic energy of the electron is 102.71 electron volts (eV).

Andre BalkonE

Result:
a) The electron moved into a region of higher potential.
b) The potential difference that stopped the electron is $34.16×{10}^{-5}$ V.
c) The initial kinetic energy of the electron is approximately $34.16×{10}^{-5}$ eV.
Solution:
a) To determine whether the electron moved into a region of higher potential or lower potential, we need to consider the effect of the electric field on the electron's motion.
When an electron moves in an electric field, it experiences a force that opposes its motion. The direction of this force depends on the charge of the electron (negative) and the direction of the electric field. If the electric field is in the same direction as the electron's motion, the force will act in the opposite direction, slowing down the electron. Conversely, if the electric field is in the opposite direction to the electron's motion, the force will act in the same direction, accelerating the electron.
In this case, since the electron is brought to rest, we can conclude that the electric field is in the same direction as the electron's initial motion. Therefore, the electron moved into a region of higher potential.
b) The potential difference, often denoted as ΔV (delta V), is the difference in electric potential between two points. It represents the change in potential energy per unit charge as a charge moves between those two points. We can use the formula:
$\Delta V=\frac{\Delta U}{q}$
where ΔU is the change in potential energy and q is the charge.
Since the electron is brought to rest, its initial kinetic energy is fully converted into potential energy. Thus, the change in potential energy is equal to the initial kinetic energy of the electron. We can express this as:

Now, substituting the given values:
$\Delta V=\frac{6.00×{10}^{5}\phantom{\rule{0.167em}{0ex}}\text{m/s}×{m}_{e}}{q}$
where ${m}_{e}$ is the mass of the electron and q is the charge of the electron.
The charge of an electron is $q=-1.6×{10}^{-19}\phantom{\rule{0.167em}{0ex}}\text{C}$, and the mass of an electron is ${m}_{e}=9.11×{10}^{-31}\phantom{\rule{0.167em}{0ex}}\text{kg}$. Substituting these values:
$\Delta V=\frac{6.00×{10}^{5}\phantom{\rule{0.167em}{0ex}}\text{m/s}×9.11×{10}^{-31}\phantom{\rule{0.167em}{0ex}}\text{kg}}{-1.6×{10}^{-19}\phantom{\rule{0.167em}{0ex}}\text{C}}$
Simplifying:
$\Delta V=\frac{-54.66×{10}^{-24}\phantom{\rule{0.167em}{0ex}}\text{kg}·{\text{m}}^{2}·{\text{C}}^{-1}}{-1.6×{10}^{-19}\phantom{\rule{0.167em}{0ex}}\text{C}}$
$\Delta V=34.16×{10}^{-5}\phantom{\rule{0.167em}{0ex}}\text{V}$
Hence, the potential difference that stopped the electron is $34.16×{10}^{-5}$ V.
c) To find the initial kinetic energy of the electron in electron volts (eV), we can use the relationship:
$1\phantom{\rule{0.167em}{0ex}}\text{eV}=1.6×{10}^{-19}\phantom{\rule{0.167em}{0ex}}\text{J}$
The initial kinetic energy of the electron is equal to the work done by the electric field to bring the electron to rest. This work can be calculated as the product of the potential difference and the charge of the electron:

Substituting the given values:

Simplifying:

To convert this value to electron volts (eV), we divide the initial kinetic energy by the conversion factor:

Since energy cannot be negative in this context, we take the absolute value:

Therefore, the initial kinetic energy of the electron is approximately $34.16×{10}^{-5}$ eV.

Do you have a similar question?