# Mass-energy relation (Einstein)

The relation between the decrease of mass and the energy that releases is given by

E = Δm.c2

• E      energy in J
• Δm  decrease in mass (mass defect)  in kg
• c       speed of light (in vacuum) in m.s -1    cvac = 2.99792458 x 108 m.s-1

Example

α-decay and the release of energy

Atomic mass U-235 = 235.038893 u

Atomic mass He-4 = 4.002603 u

Atomic mass Th-231 = 231.03629 u

u  : atomic mass unit

1 u = 1.66054 x 10-27 kg

We compare the mass before en after the disintegration

mass U-235 :  235.04393 u                           mass   He-4 (α-particle) :      4.002603 u

mass   Th-231                  : 231.03629   u     _

____________________

235.038893 u

Mass decreases with : 235.038893 u – 235.04393 u  = 0.005037 u  (mass defect)

Released Energy

E = Δm.c2 =( 0.005037 x 1.66054 x 10-27) (2.99792458 x 108)2 =  7.539315488 x 10-13   J =

4.7121 x 106 eV

Also possible:

E = 0.005037 x 931.49 = 4.6919 MeV

(1 u is equivalent with 931.49 MeV)

Remark

An atomic mass includes the mass of the orbital electrons.  This causes no error, because the number of electrons included by U-235 is equal to the number of electrons of He-4 and Th-231.

Example

β–   decay and the release of energy

Before disintegration:

mass P-32 atom :                                            31.97391 u

mass 15 electrons   : 15 x 0,00055 u =          0.00825 u   _

—————————-

mass P-32 nucleus                                         31.96566  u

After disintegration

massa S-32 atom :                                              31.97207 u

massa 16 electrons  : 16 x 0,00055 u=          0.0088   u          _

—————————–

massa S-16 nucleus                                          31.96327 u

massa   β ( electron)                                        0.00055 u

Mass decreases with :

Δm =31.96566 u – ( 0.00055 u + 31.96327 u) = 31.96566 u – 31.96382 u = 0.00184 u

Released Energy

E = Δm.c2 =(0.00184 x 1.66054 x 10-27)  ( 2.99792458 x 108)2 = 2.74605 x 10-13 J = 1.7163 x 106 eV =

1.7163 MeV 