How Do You Convert Volts to Joules for Calculating Electron Velocity?

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SUMMARY

The discussion focuses on converting volts to joules to calculate the maximum velocity of an electron accelerated by a potential difference of 10,000 volts. The conversion is established as 1 electron-volt (eV) equaling approximately 1.60217646 x 10^-19 joules. The kinetic energy formula used is KE = 0.5 x M x V^2, where M is the mass of the electron. Participants emphasize the importance of precision in calculations and avoiding premature rounding to minimize errors.

PREREQUISITES
  • Understanding of kinetic energy formula (KE = 0.5 x M x V^2)
  • Knowledge of electron mass and charge
  • Familiarity with the concept of electron-volts (eV)
  • Basic principles of electricity and energy conversion
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  • Research the mass of an electron (9.109 x 10^-31 kg)
  • Learn about energy conversion between electron-volts and joules
  • Study the effects of rounding in scientific calculations
  • Explore advanced kinetic energy calculations in particle physics
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Students in physics, electrical engineers, and anyone interested in particle acceleration and energy conversion principles.

icedevilwoot
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Hey.
I just got up to this question which I am stuck on an need a bit of help.
Electron is accelerated by a potential difference of 10,000 volts
1) What is the max velocity it will attain
Now, I tried using the ke = 0.5 x M x V^2
Just how do I convert Volts to joules and is that the correct formula I am using? Thanks.
 
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You need to convert "electron-volts" (eV) to joules. The eV is the energy of one electron accelerated across one volt. To get to joules, you use the same number as the number of electrons in one coulomb (since a joule is a "coulomb-volt"). You will get a very small number for joules. THen proceed with the KE equation.
 
<br /> 1 eV = 1.6 x 10^-19 Joules<br />
 
Dr.Brain said:
<br /> 1 eV = 1.6 x 10^-19 Joules<br />
didn't you mean 1 eV \approx 1.6 \times 10^-19 Joules
or to better precision 1 eV \approx 1.60217646 \times 10^-19 Joules
 
Stop being so critical.

1 eV \approx 1.60217646 \times 10^{-19} so youre both wrong.
 
Ouabache said:
didn't you mean 1 eV \approx 1.6 \times 10^-19 Joules
or to better precision 1 eV \approx 1.60217646 \times 10^-19 Joules

Yes, that was what he meant- that was exactly what he said.

And, of course, since the data was given to two significant figures, so was his answer!
 
HallsofIvy said:
Yes, that was what he meant- that was exactly what he said.

And, of course, since the data was given to two significant figures, so was his answer!
Wanted to indicate to the questioner, that Doc's value is not absolute. :rolleyes:
As you correctly alluded, it's rounded off.

In my own experience, I've found it better not to round off until reaching the final answer. Even in academic questions, I've often encountered solution discrepancies due to accumulation of rounding errors.

I wish to apologize to icedevilwoot on behalf of us (mentors) for this lengthy digression.
I wonder how are you making out with your calculations? :smile:
 
Last edited:

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