Point A is at a potential of +250 V, and point B is at a potential of -150 V. An á-particle is a helium nucleus that contains two protons and two neutrons; the neutrons are electrically neutral. An á-particle starts from rest at A and accelerates toward B. When the á-particle arrives at B, what kinetic energy (in electron volts) does it have?

Question

Point A is at a potential of +250 V, and point B is at a potential of -150 V. An á-particle is a helium nucleus that contains two protons and two neutrons; the neutrons are electrically neutral. An á-particle starts from rest at A and accelerates toward B. When the á-particle arrives at B, what kinetic energy (in electron volts) does it have?

Answer 1

The charge is 2e and the voltage change is 400 V. Multiply them together for the kinetic energy gain.

You could also compute the final speed using the alpha particle mass, but they don't ask for that.

Answer 2

To find the kinetic energy of the α-particle at point B, we need to calculate the change in potential energy.

The potential energy of a charged particle in an electric field is given by the formula:

PE = qV

Where:
PE = Potential Energy
q = Charge of the particle
V = Potential difference

In this case, the charge of the α-particle is +2e (since it consists of two protons), and the potential difference (V) is the difference in potential between point A and point B.

Given:
Potential at point A = +250 V
Potential at point B = -150 V

The potential difference between A and B (ΔV) is given by:

ΔV = Vb - Va

Substituting the values:

ΔV = -150 V - (+250 V)
= -400 V

Now we can calculate the change in potential energy:

ΔPE = (2e) * ΔV

Substituting the charge of the α-particle (2e) and the potential difference (-400 V):

ΔPE = (2 * 1.6 x 10^-19 C) * (-400 V)
= -1.28 x 10^-18 J

To convert this energy to electron volts (eV), we use the conversion factor:

1 eV = 1.6 x 10^-19 J

So, the change in potential energy in electron volts is:

ΔPE(eV) = -1.28 x 10^-18 J / (1.6 x 10^-19 J/eV)
≈ -8 eV

Since the α-particle loses potential energy as it moves from A to B, the change in potential energy ΔPE will be negative. Therefore, the kinetic energy of the α-particle at point B is positive 8 eV.

Answer 3

To find the kinetic energy of the α-particle when it arrives at point B, we need to first find the change in potential energy that it experiences as it moves from point A to point B, and then convert that into kinetic energy.

The change in potential energy (ΔPE) can be found using the formula:

ΔPE = qΔV

Where q is the charge and ΔV is the change in voltage.

In this case, the α-particle has a charge of +2e (where e is the elementary charge), since it contains two protons. So:

ΔPE = (2e)(ΔV)

Now, we need to calculate ΔV, which is the difference in voltage between point A and point B:

ΔV = Vb - Va
ΔV = (-150 V) - (+250 V)
ΔV = -400 V

Plugging this value into the formula for ΔPE, we get:

ΔPE = (2e)(-400 V)
ΔPE = -800eV

Negative sign indicates a decrease in potential energy.

Finally, since the potential energy at point B is converted into kinetic energy (KE) when the α-particle arrives there, we can say that:

KE = -ΔPE
KE = -(-800eV)
KE = 800eV

So, the kinetic energy of the α-particle when it arrives at point B is 800 electron volts (eV).