FE Physics
Formulas & Explanation

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 1

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 2

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

Binding energy

The binding energy of a nucleus is the energy that would be needed to split up the nucleus into separate nucleons .

Example

We calculate the binding energy of the He-4 nucleus

This nucleus consists of 2 protons en 2 neutrons

Mass  of the 4 particles    : 2 x 1.007276 u + 2 x 1.008665 u = 4.031882 u

Mass of the nucleus          :  4.002603 u – 2 x 0.00054858 u = 4.00150584 u

Mass defect Δm                  :  0.03037616 u

Binding energy                    :  E= Δm.c2 =  (0.03037616 x 1.66054 x10-27) (2.99792458 x 108)2 = 4.546664 x 10-12 J = 2.8417 x 107 eV = 28.42 MeV

Binding energy  per nucleon = 7.10 MeV

Remark : The smaller the mass per nucleon the more stable the nucleus.