Point A is at a potential of +220 V, and point B is at a potential of -140 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?
To determine the kinetic energy (KE) of the α-particle when it arrives at point B, we need to consider the change in electric potential energy (PE) and the conservation of energy principle.
Here's how you can find the kinetic energy:
1. Determine the change in electric potential energy:
The change in electric potential energy (ΔPE) is equal to the product of the charge (q) and the change in electric potential (ΔV).
ΔPE = q * ΔV
In this case, since the α-particle has a charge of +2e (where e is the elementary charge), and the difference in potential is given as -140V - 220V = -360V:
ΔPE = (2e) * (-360V)
= -720eV
Note that the negative sign indicates a decrease in potential energy as the α-particle moves from A to B.
2. Apply the conservation of energy principle:
According to the conservation of energy, the change in potential energy (ΔPE) is converted into kinetic energy (KE). Therefore:
ΔPE = KE
Substituting the value of ΔPE from step 1:
KE = -720eV
Since 1 electron volt (eV) is defined as the amount of kinetic energy gained or lost by an electron when it passes through a potential difference of 1 volt, we can write:
1eV = 1.6 x 10^-19 J (the conversion factor)
Multiplying the conversion factor to express the result in joules:
KE = -720eV * (1.6 x 10^-19 J/eV)
≈ -1.15 x 10^-16 J
Since we want to find the kinetic energy in electron volts, we need to convert the kinetic energy back to electron volts:
KE (eV) = -1.15 x 10^-16 J / (1.6 x 10^-19 J/eV)
≈ -7.2 x 10^2 eV
As kinetic energy cannot be negative, the magnitude of the kinetic energy of the α-particle when it arrives at point B is approximately 720 electron volts (eV).