16.2. Electric Potential http://www.ck12.org
It is simplest to think that for every one volt of potential difference the particle experiences, it gains (or loses) 1eV.
- An alpha-particle (the nucleus of a helium atom) is fired toward the positive plate of a parallel plate conductor
and passes through a potential difference of 1,500 V. What is the change in its kinetic energy? Express your answer
in eV.
Answer:
Protons are the only charges inside the nucleus of an atom and so the alpha particle must be positively charged. A
helium nucleus contains two protons (and two neutrons) with a total charge of 2( 1. 60 × 10 −^19 C).
The alpha particle must slow down due to the electrostatic repulsion from the positive plate. It must, therefore, lose
kinetic energy and gain potential energy.
Each proton loses 1,500eVof kinetic energy.∆KE=− 3 , 000 eV.
- An electron and a proton both gain kinetic energy of 1eV.
True or False: Their speeds must be the same, since they both gained the same amount of energy.
Answer: False. The mass of a proton is nearly 2000 times greater than the mass of an electron. Remember that
kinetic energy depends on both the speed and mass of an object. Therefore, the final speed of the electron will be
much greater.
Illustrative Example 16.2.5
a. An electron and a proton both gain kinetic energy of 1 eV. What is the ratio of the electron’s speed to the proton’s
speed?
Answer:
As discussed above, though both particles gain the same kinetic energy, their speeds will not be the same, since they
have different masses. The mass of the proton is nearly 2000 times as great as the electron’s so:
KEe
KEp=
1
2 mev2
e
1
2 mpv(^2) p=^1 →mev
2
e=mpv
2
p→
v^2 e
v^2 p
=
mp
me=
2000 me
me= 2 , 000 →
ve
vp=
√
2 , 000 = 44. 7 → 45
The electron will move about 45 times faster than the proton.
b. What is the speed of a proton which has a kinetic energy of 37MeV?
The mass of a proton is 1. 67 × 10 −^27 kg.
Answer:
Because the electron-volts are a very small unit, they are typically expressed inKeV(1000 electron-volts) andMeV
(one million electron-volts). The electron-volt is a convenient unit of measure but it is not an SI unit. In order to find
the velocity of a particle if its energy is given in units ofeV, we must convert back into Joules.
37 MeV= ( 37 × 106 )( 1. 60 × 10 −^19 J) = 5. 92 × 10 −^14 J
1
2
mpv^2 = 5. 92 × 10 −^14 J→1
2
( 1. 67 × 10 −^27 kg)v^2 = 5. 92 × 10 −^14 J→v= 8. 4 × 106
m
s