Convert atomic mass into Joules

[Sebastiaan]

Homework Statement

How much energy is created when 0,0188828901 atomic mass is converted into energy (Joule)

Homework Equations

E = m * c^2
speed of light = 299792458
1 atomic mass unit = 1.660539040e-27 kg
1 atomic mass unit = 931,494028 MeV
1 eV = 1,6021766208e-19 Joule

The Attempt at a Solution

[/B]
mass = 0,01888289012 * 1,660539040e-27 = 3,13557762322902848e-29

Applying this to E= mc^2 I get

3,13557762322902848e-29 * 299792458 * 299792458= 2,818116627e-12 joules

Another method is that I first convert mass to MeV and then to Joules I get

0,01888289012 * 931,494028 * 10e6 * 1,6021766208e-19 = 2,8181164240e-12 joules

However when I attempt to atomic mass to Joule using a online converter

(which uses 1 unified atomic mass unit = 1.49241795273396E-10 joule ) I get

0,01888289012 * 1,49241795273396e-10 = 2,818116421e-12 joules

As you can see I get different outcomes depending on the used method.

Which one of the result would be most accurate?

 

[BvU]

From your link:

The unified atomic mass unit u or dalton Da is the standard unit that is used for indicating mass on an atomic or molecular scale. It is defined as one twelfth of the mass of a carbon-12 atom or 1.6605•10⁻²⁷ kg. During a nuclear reaction the mass of its products is less than the mass of the initial substance. This is so because the mass is converted into energy. The energy liberated in the nuclear reaction is given by Einstein's equation E=mc² where E is the energy, m is the mass and c = 299 792 458 m/s is the speed of light. Therefore, (1 J)/c² = 6.700 535 85(30)•10⁹ u
​

I see 8 digits in 1J/c², so 1.49241795273396 10^-10 suggests too many digits.

For fundamental constants, consult the http://pdg.lbl.gov/2017/reviews/rpp2016-rev-phys-constants.pdf results

 

[Sebastiaan]

For fundamental constants, consult the http://pdg.lbl.gov/2017/reviews/rpp2016-rev-phys-constants.pdf results

If I'm not mistaken, It does not appear to list atomic mass to Joule.

 

[BvU]

$c$ is exact

 

[Sebastiaan]

Yes but the 1 eV = 1.672621898e-27 kg is only 10 digits from which the last 2 digits are uncertain with a mean deviation of 0.0000000021 so only 7 digits are truly accurate

 

[BvU]

I get 2.818116624(34)E-12 J

 

[BvU]

Don't take the proton mass; take the amu !

 

[Sebastiaan]

I get 2.818116624(34)E-12 J

I'm confused, how exactly did you calculate this?

 

[BvU]

0.0188828901000 * 1.66053904E-27 * 299792458^2 = 2.81811662422E-12 in excel (works in double precision, so some 15 digits; good enough here)

http://pdg.lbl.gov/2017/reviews/rpp2016-rev-phys-constants.pdf in 1.660539040E-27 is 2E-35, so relative unertainty 1.2E-8

Result times relative uncertainty gives 3.4E-20 hence reported as 2.818116624(34)E-12 J