# Mass-Energy Equivalence

1. May 27, 2015

### says

1. The problem statement, all variables and given/known data
Calculate the energy equivalent of the mass of an alpha particle.

alpha particle (amu) = 4.00150618 u
1 amu = 1.660566*10-27 kg
1 amu = 931.5020 MeV
alpha particle (kg) = 6.64465675 * 10-27 kg

2. Relevant equations
e = mc2

3. The attempt at a solution
931.5020 MeV * 4.00150618 u
= 3727.41100968 MeV (i)
= 3727.411
(correct to 7 significant figures because 931.5020 MeV has 7 significant figures)

6.64465675 * 10-27 kg * ((2.99792458 * 108 m/s)2) = 5.97191966 * 10-10 Joules

1 eV = 1.602189 * 10-19 J

5.97191966 * 10-10 Joules / 1.602189 * 10-19 Joules

= 3727350306.36 eV
= 3727.35031 MeV
= 3727.350 MeV (correct to 7 significant figures because 1.660566*10-27 kg has 7 significant figures)

I don't understand which answer is correct. I know how to calculate the mass-energy equivalence but I'm stumped on precision & accuracy.

2. May 27, 2015

### BvU

Hi there,

Strange, since a simple google shows for the alpha particle

6.64465675(29)×10−27 kg[1]
4.001506179125(62) u
3.727379240(82) GeV/c2​

and for your conversion factors the claims also don't hold (e.g. pdg or nist):

u = 931.494061(21) MeV/c2
u = 1.660538921(73) × 10−27 kg
numbers between brackets are uncertainties. If you want to dive really deep, you have to take into account that these uncertainties are correlated and you have to go back to the (84 page) original article . Wouldn't do that. Just keep in mind that not all physical constants are known to arbitrary precision.

3. May 27, 2015

### says

My question is more about significant figures, accuracy and precision though.

1 amu = 931.5020 MeV
alpha particle = 4.00150618 u

energy equivalent of alpha particle = 3727.41100968 MeV

931.5020 only has 7 significant figures though, so doesn't that mean the energy equivalent = 3727.411 MeV?

If this is the case, then what about the other calculation I did?

6.64465675 * 10-27 kg * ((2.99792458 * 108 m/s)2) = 5.97191966 * 10-10 Joules

5.97191966 * 10-10 Joules = energy equivalent of alpha particle (in Joules)

1 eV = 1.602189 * 10-19 J

5.97191966 * 10-10 Joules / 1.602189 * 10-19 Joules

= 3727350306.36 eV
= 3727.35031 MeV
= 3727.350 MeV (correct to 7 significant figures because 1.660566*10-27 kg has 7 significant figures)

I don't know which starting value I should base my significant figures in the final answer on.

4. May 27, 2015

### BvU

I'm telling you 1 amu = 931.494061(21) MeV/c2 and I give you the reference where it comes from.

How can you say your 931.5020 value is accurate to 7 digits ? Says who ? Where does it come from ?

4.001506179125 * 931.494061 = 3727.379241
4.00150618 * 931.4940610 = 3727.3792417​

both well within the range of 3727.379240(82) MeV/c2
but

4.00150618 * 931.5020000 = 3727.4110097​

is "way off" when considering the 82 eV uncertainty in the pdg value..

And the same for the other path:

6.64465675(29) 10-27 is what I found too
2.99792458 108 is exact, so
5.971919665 10-10 is what I found too , but
1.602176565(35) 10-19 is what I found for e -- so you are off by a factor 1 + 7.8 10-6
Where does your 1.602189 come from ?

1.602176565 gives 3727.379239
again well within the range of 3727.379240(82) MeV/c2
but

1.602189 gives 3727.350309​

and that is again "well off".

Fitting the fundamental physical 'constants' is an elaborate business: they have to evaluate a huge load of experimental results and then do a least squares fit.
Out comes a set of values (and corrleated sigmas) with maximal self-consistency. See the CODATA paper.

Could it be that in your case data from an old set is compared to results that are based on the newest values ?
You'd have to dive into the archives.

--

There is something else that doesn't match. You say
4.00150618 * 1.660566 = 6.64465675

but it's not. It's 6.64476511
so that is off too. And the 7 digit accuracy of the u in kg has nothing to do with the deviation; it's all due to the "wrong" e.

Last edited: May 27, 2015