When is it safe to use eV units and when do we have to convert eV units into kg?

  • Thread starter Thread starter DunWorry
  • Start date Start date
  • Tags Tags
    Convert Ev Units
Click For Summary
SUMMARY

This discussion clarifies the conditions under which electronvolts (eV) can be used in physical formulas. In the formula p = √((E² - m²c⁴)/c²), energy and mass can be expressed in MeV, yielding a valid result. However, in the formula e^(-√((2m(V-E))/h²)x), using eV for mass, potential, and energy leads to incorrect results unless mass is converted to kilograms and energy to joules. The key takeaway is that when mass is involved in a formula that also includes spatial dimensions, all units must be consistent, necessitating conversions to maintain dimensional integrity.

PREREQUISITES
  • Understanding of relativistic energy-momentum relationships
  • Familiarity with quantum mechanics and wave functions
  • Knowledge of unit conversions between eV, joules, and kilograms
  • Basic grasp of dimensional analysis in physics
NEXT STEPS
  • Study the implications of dimensional analysis in physics problems
  • Learn about unit conversions specifically between eV and joules
  • Explore the role of Planck's constant (ℏ) in quantum mechanics
  • Investigate additional examples of energy-mass relationships in relativistic physics
USEFUL FOR

Students of physics, particularly those studying quantum mechanics and relativity, as well as educators looking to clarify unit conversion principles in energy calculations.

DunWorry
Messages
40
Reaction score
0

Homework Statement


I have a misunderstanding of when I can safely use eV units in a formula. For example in the formula p = \sqrt{\frac{E^{2} - m^{2}c^{4}}{c^{2}}} I can put the energy and mass in terms of MeV and get an answer with units MeV/C, which makes sense. But then there was this formula I was using e^{-\sqrt{\frac{2m (V-E)}{h^{2}}}x}

where m is mass, v is potential, e is energy, h bar squared on bottom (I couldn't find symbol) and x is in metres. It is supposed to give probability which is dimensionless. However, if I put in mass, potential and energy in terms of eV, I get a wrong answer. if I convert the mass into kg, put potential and energy in terms of joules then I get the correct answer. Why could I use eV in the first case but had to convert in the second case? is it because I am multiplying by x which is in metres so it somehow does not work with eV? what if I had another formula with mass/energy and was multiplying by a speed with units m/s? would I again have to convert all eV units into kg/joules etc?

Are there any general rules I should keep in mind whilst working with eV?

Thanks
 
Physics news on Phys.org
Welcome to PF!

Hi DunWorry! Welcome to PF! :smile:

(it's the obvious … \hbar :wink:)

The only unit you've changed is the mass unit (by a factor MeV/kg).

If the terms in the top line of your formula all contain MeV, then since the bottom line (c2) doesn't contain mass at all, there's no difficulty: the result will be in MeV.

In your second formula, the bottom line (##\hbar##2) does contain mass, so it won't work unless you rewrite h in terms of Mev. :wink:

(so the general rule is that it's ok to change the mass unit, so long as you change it everywhere)​
 

Similar threads

Replies
8
Views
5K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
4
Views
5K
  • · Replies 8 ·
Replies
8
Views
5K
Replies
0
Views
1K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 25 ·
Replies
25
Views
4K