Contents
Advanced Physics Topics like quantum mechanics and relativity have revolutionized our understanding of the universe.
What is Meant by Mass Excess? What is the Formula of Nuclear Density?
Definition: The energy that keeps protons and neutrons confined to the nucleus, is called nuclear binding energy.
If an amount of energy equal to the nuclear binding energy is supplied from outside then the nucleus disintegrates and the protons and neutrons exist as free particles. Hence, binding energy of a nucleus is also defined as the external energy required to separate the constituents of the nucleus.
Relation between binding energy and mass defect: Binding energy of a nucleus can be explained using mass-energy equivalence. When protons and neutrons exist freely, sum of their individual masses gives the ‘mass energy’ of the system. But when these very protons and neutrons form a nucleus, both nuclear binding energy and a nuclear mass exist.
Hence, from the law of conservation of mass-energy,
(sum of masses of protons and neutrons) × c2 = mass of nucleus × c2 + nuclear binding energy
If Z = atomic number and A = mass number of the nucleus and mp and mn, mass of proton and mass of neutron independently, the conservation condition can be mathematically expressed as
Zmpc2c2 + (A – Z)mnc2 = \(M_{Z, A} c^2\) + ∆E
where MZ, A = mass of nucleus and ∆E = binding energy
Hence, ∆E = {Zmp + (A – Z)mn – MZ, A}c2 ……. (1)
This can be written as
∆E = ∆m ᐧ c2 …….. (2)
The expression within second bracket in equation (1) represents ∆m. When a nucleus is formed from its nucleons, the mass of the nucleus is less than the masses of the nucleons taken together. This means that {Zmp + (A – Z)mn} is greater than MZ, A. This difference Is called mass defect, ∆m.
In fact, this reduced mass is transformed into binding energy to form nucleus.
Example: Mass of proton mp = 1.0073 u and that of neutron mn = 1.0087 u. Since nucleus of He4 consists of 2 protons and 2 neutrons, their total mass
= 2 × mp + 2 × mn = 2 × 1.0073 + 2 × 1.0087 = 4.0320 u
Experimental value of mass of He4 nucleus, M2, 4 = 4.0015.
Hence, mass defect in He4 nucleus
∆m = [2mp + 2mn] – M2, 4 = 4.0320 – 4.0015 = 0.0305 u
∴ Binding energy of He4 nucleus
∆E = 0.0305 × 931.2 MeV = 28.4016 MeV
Hence, binding energy per nucleon or average binding energy of He4 nucleus = \(\frac{\Delta E}{4}\) = \(\frac{28.4016}{4}\) MeV = 7.1004 MeV
Stability of a nucleus: More the binding energy of a nucleus, more is the energy required to separate its nucleons and hence the nucleus is more stable. In this respect a nucleus can be compared with a liquid drop. A very large liquid drop has a tendency to break up into smaller drops whereas a large number of small drops tend to join to form a large drop. Similarly, a large nucleus (like that of U238) and small nucleus (like H2) both are unstable.
Stability of a nucleus depends upon the binding energy per nucleon. The graph of binding energy per nucleon of different elements vs mass number is given in Fig. .
From the graph we observe that when each nucleon acquires an average binding energy of 8MeV, the nucleus becomes most stable. As seen earlier, for He4 nucleus, the value is 7.1 MeV and therefore He4 is a stable nucleus. Maximum value of binding energy per nucleon is observed for Ni62 which is 8.8 MeV/nucleon. Majority of the nuclei, as shown in the graph, lying within the range A = 12 and A = 206 are stable nuclei.
Significance of the binding energy curve:
i) The nearly constant value of the binding energy per nucleon for nuclei of about A = 25 to A = 170 shows a saturation, since no further increase occurs if extra nucleons are added.
ii) The saturation means that nuclear force is a short range force-any extra nucleon, when added, resides on the surface, having no effect on the nucleons deep inside the pre-existing nucleus.
iii) The binding energy per nucleon for A higher than about 220 is less than that at the middle region of the periodic table. So a heavy nucleus tends to break up into two mid-range nuclei to attain higher stability—this is nuclear fission. A significant amount of energy is released in this process.
iv) Nuclei with say, A < 4 to 6 have a relatively low value of binding energy per nucleon. So two or more of them tend to unite into one nucleus with a higher value of binding energy per nucleon, to attain stability. This is nuclear fusion. In this process also, a significant amount of energy is released.
Mass excess: For C12 atom, A = 12 and in this case, according to the definition of atomic mass the actual mass of C12 atom, M = 12 u. However, for all other elements the value of A and M are different. For example, for He4, A = 4 but M = 4.002603 u, for O16, A = 16 but M = 15.994915 u. This difference in the value A and experimental value of M is called mass excess. Clearly mass excess can be either positive or nega-tive for different nuclei.
Definition: If mass number = A and atomic mass = M of a nuclide then mass excess of that nucleus, ∆M = M – A.
From the above examples, mass excess of the elements are listed below.
Elements | Mass excess |
(i) C12 | 0 |
(ii) He4 | 0.002603 u |
(iii) O16 | -0.005085 u |
Volume and Density of a Nucleus
Different nuclei are similar to a drop of liquid of constant density. The volume of a liquid drop is proportional to its mass, which is proportional to the number of molecules contained in it. Similarly, the nuclear density is also a constant quantity. So the nuclear volume is directly proportional to the mass number and is independent of the separate values of the proton number and the neutron number.
Radius of nucleus: Experimentally it has been found that a proton or a neutron has a radius, r0 = 1.2 × 10-13 cm = 1.2 × 10-15m
So the volume of each proton or neutron, V’ = \(\frac{4}{3} \pi r_0^3\)
Let the total number of protons and neutrons in the nucleus = mass number = A
Then the volume of the nucleus, V = \(\frac{4}{3} \pi r_0^3 A\)
If R be the radius of the nucleus, then V = \(\frac{4}{3} \pi R^3\)
Hence, R3 = \(r_0^3 A\) or, R = r0A1/3
For A = 216, we get R = 6r0 = 7.2 × 10-13cm.
Hence even the radius of a heavy nucleus is less than 10-12cm.
Calculation of nuclear density: Estimated mass M of a nucleus of mass number A,
M = A u = A × 1.66 × 10-24g
Also, volume of this nucleus, V = \(\frac{4}{3} \pi r_0^3\) × A
Hence, nuclear density,
ρN = \(\frac{M}{V}\) = \(\frac{A \times 1.66 \times 10^{-24}}{\frac{4}{3} \pi r_0^3 \cdot A}\)
= \(\frac{3 \times 1.66 \times 10^{-24}}{4 \pi \times\left(1.2 \times 10^{-13}\right)^3}\) [r0 = 1.2 × 10-13cm]
= 2.3 × 1014 g ᐧ cm-3
Generally, nuclear density is taken as 2 × 1014 g ᐧ cm-3 or
2 × 1017 kg ᐧ m-3, which is very high and represents the presence of a lot of mass concentrated within a very small space. So, nuclear density is more than 1014 times the density of water.
Numerical Examples
Example 1.
For a nucleus which is nearly spherical in shape r = r0A1/3, where r is the radius and A is the mass number and r0 is a constant of value 1.2 × 10-15 m. If the mass of the neutrons and protons are equal and equal to 1.67 × 10-27 kg, prove that density of the nucleus is 1014 times the density of water.
Solution:
Mass of A -number of neutrons and protons ≈ mass of nucleus (M) = 1.67 × 10-27 × A kg
Again, volume of the nucleus,
V = \(\frac{4}{3}\)πr3 = \(\frac{4}{3} \pi r_0^3 A\)
= \(\frac{4}{3}\) × 3.14 × (1.2 × 10-15)3 ᐧ A m3
∴ Density of nucleus = \(\frac{M}{V}\) = \(\frac{1.67 \times 10^{-27} \times A}{\frac{4}{3} \times 3.14 \times\left(1.2 \times 10^{-15}\right)^3 \cdot A}\)
= 2.3 × 1017 kg ᐧ m-3
∴ \(\frac{\text { density of nucleus }}{\text { density of water }}\) = \(\frac{2.3 \times 10^{17}}{1000}\) = 2.3 × 1014
∴ Density of nucleus is more than 1014 times the density of water. (Proved)