A helium ion with +2 elementary charges is accelerated by a potential difference of 5.0 x 10^3 volts. What is the kinetic energy acquired by the ion?

Asked on by acaputi

1 Answer | Add Yours

Top Answer

txmedteach's profile pic

txmedteach | High School Teacher | (Level 3) Associate Educator

Posted on

This problem is likely to be looking for a simple solution. You can actually see the simplicity in the units! The problem is asking for an answer in electron-volts, which is simply in terms of an elementary charge multiplied by volts.

The electron is the elementary charge. Almost everything in particle physics is done in electron-volts (eV) and, as such, relies on the charge of the electron as the elementary unit. In our case, the helium ion is stated to have a charge of +2 elementary charges. This statement basically means it is +2 times the charge of an electron!

Now, we know the change is being accelerated by a potential difference of 5.0*10^3 volts. There is no information about distance, time, or any other factors, so we will just assume that the helium ion is travelling some distance that it will get basically the full effect of the voltage.

What does this mean? Look at the units! We are just going to multiply the charge by the voltage! Another way of thinking about it? The volt is a unit of energy/charge. If we multiply by charge, the charge part of it cancels out and we are left with energy!

So, for the answer, all you do is multiply +2 e (charge of an electron) with the voltage:

+2 e * 5.0*10^3 V = 10.0*10^3 eV

Scientists are also fans of putting letters for every factor of 1,000, like the one we have here. This number can then be converted in the following way:

10.0*10^3 eV = 10.0 keV

Either one of these can work as your answer! Remember, if you get lost on these sorts of problems, 9 times out of 10, you just need to multiply and divide in ways that make the units what you want! This method is called "dimensional analysis."

I hope that helps!


We’ve answered 319,862 questions. We can answer yours, too.

Ask a question