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

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Using eV units in formulas is acceptable when all terms are consistently expressed in the same energy unit, such as MeV. In the first formula, energy and mass in MeV yield a coherent result since the denominator does not involve mass. However, in the second formula, the presence of mass in the denominator necessitates converting all quantities to compatible units, such as joules, to ensure dimensional consistency. This inconsistency arises because the formula involves a multiplication by a distance in meters, which requires energy to be in joules. A general rule is to maintain uniformity in units across all terms in a formula to avoid errors.
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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
 
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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)​
 
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