Question

A common unit used to measure the energy of small particles is the electron volt (eVeV), which is equal to the magnitude of the charge on the electron (measured in coulombs CC) multiplied by 1 volt (VV). In chapter 17, you will learn that coulombs multiplied by volts is equivalent to mass multiplied by length squared and divided by time squared.

Which of the following quantities could be written in eVeV ?
Case A: gravitational potential energy m????ℎmgh, where mm is the mass of the object, ????g is the acceleration due to gravity (length divided by time squared), and ℎh is the height of the object measured with respect to a reference position
Case B: electric potential energy ????Δ????qΔV , where ????q is the charge of the particle in coulombs and Δ????ΔV is the potential difference in volts through which the particle is being accelerated
Case C: rest energy mc2mc2, where mm is the mass of the object, and cc is the speed of light (length divided by time)
Case A and Case B
Case C
all of them
Case B
Case A and Case C
Case A
Case B and Case C
Explain your reasoning:

Answer 1

ANSWER -

The correct answer is "Case B". The electric potential energy, which is ????Δ????qΔV , can be written in eVeV because it is equal to the charge of the particle in coulombs multiplied by the potential difference in volts. The electron volt is a measure of the energy of small particles, so this quantity is appropriate to express in eVeV.

The gravitational potential energy, m????ℎmgh, cannot be expressed in eVeV because it is not related to the electric charge of a particle and is not in the appropriate units.

The rest energy, mc2, cannot be expressed in eVeV because it is a measure of the energy of a particle at rest, which does not involve acceleration through a potential difference.

So, only Case B can be expressed in eVeV.