Question

Point A is at a potential of +250 V, and point B is at a potential of -150 V. An α-particle is a helium nucleus that contains two protons

and two neutrons; the neutrons are electrically neutral. An α-particle starts from rest at A and accelerates

toward B. When the α-particle arrives at B, what kinetic energy (in electron volts) does it have?

Answer 1

The charge of the alpha particle is equal to the charge of two electrons (by modulus):

$$q = 2|e|$$

where $e = -1.6\times10^{-19}C$ is the charge of electron.

By definition, the potential difference between two points $A$ and $B$ is equal to the work that should be done to move a unit positive charge from $A$ to $B$. Then the work required to move not unit charge $q$ is:

$$W = Vq = 2V|e|$$

where $V =250V - (-150V) = 400V$ is the absolute potential difference between two points.

According to the work-energy theorem, this work is equal to the change in the kinetic energy of the particle:

$$K_f - K_i = W$$

Since $K_i = 0$ is the initial kinetic energy of the particle, the final kinetic energy is:

$$K_f = W = 2V|e|$$

By definition, one electronvolt is the amount of kinetic energy gained by a single electron accelerating from rest through an electric potential difference of one volt in vacuum. Thus, dividing $K_f$ by the charge of the electron $|e|$, obtain the kinetic energy in electronvolts:

$$\dfrac{K_f}{|e|} = 2V = 2\cdot 400 = 800eV$$

Answer. 800eV.