This is probably a stupid question, but how do you covert a potential difference measured in volts to electron volts? Do you just multiply the volts by e= 1.60e-19 C?
I was looking at a http://chemlab.pc.maricopa.edu/periodic/isotopes.html and was wondering why the half-life was sometimes replaced with electron volts? There's an example from the table below:
Helium-6 806.7 milliseconds -> Lithium-6
Helium-7 160 KEV -> Helium-6...
current is 5 amperes. voltage is 10 volts.
1. what would be the charge if there is energy of 30 J?
2. what would be the time taken to supply the use of 30J?
For 1, is it 30J/ 10V = 3 coulombs
For 2, is it 3C/ 5 amperes = 0.60 seconds
Anyone can help?