Inorganic and Applied Chemistry

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Inorganic and Applied Chemistry

Example 1- D:

Mass and energy (Einstein equation)

From a thermodynamic point of view the stability of an atomic nucleus means that in terms of energy it is

favourable for the nucleus to exist as a whole nucleus rather than split into two parts or (hypothetically

thinking) exist as individual neutrons and protons. The thermodynamic stability of a nucleus can be

calculated as the change in potential energy when individual neutrons and protons join and form a nucleus.

As an example we are going to look at the tin isotope tin-118. Tin is element number 50 and thus this

isotope contains 50 protons and 118 – 50 = 68 neutrons in the nucleus. In order to calculate the change in

energy when the nucleus is “formed” we first have to determine the change in mass when the following

hypothetic reaction occurs:

5011 p  68 01 n 11850 Sn

The mass on the right side of this reaction is actually not the same at the mass on the left side. First we will

look at the masses and change in mass:

Mass on left side of the reaction:

Mass 5011 p 6801 n   501. 6726210

27 kg 681. 6749710

27 kg   1. 9752610

25 kg

Mass on right side of the reaction:

 kg

mol

kg mol

Mass Sn 23 1 25

118 3

(^506). 022 10 1.^9578510
117. 9016010 /
Change in mass when reaction occurs (tin-118 formation):
Masschange 1. 9578510
25 kg

1. 9578510
   25 kg


2. 7414510
   27 kg
   It is thus seen that when the reaction occurs and the tin-118 nucleus is formed, mass ”disappears”. This
   change in mass can be inserted into the famous Einstein equation (equation (1- 3) on page 18) and the
   change in potential energy can be calculated.
   E kg m s J
   E m c
   27 8 2 10
   2
   


3. 7414510
   310 /


4. 610

  
It is seen that the “disappeared” mass has been converted into 1.6 10 -10 Joules which then are released.
This corresponds to 980 MeV (1 Mega electron Volt corresponds to 1.60 10 -13 J). This amount of energy
can be translated into an amount of energy pr. nucleon:
Atoms