liquid drop model

liquid drop model

Liquid drop model of the nucleus was purposed by Neil Bohr in 1937 using the concept of potential barrier developed by gamov.

According to this model, the drop of liquid having constant charge density but of varying mass is identical to the nuclei of all elements. The assumptions of liquid drop model are based on the similarities between the liquid drop and atomic nucleus.

Assumptions of liquid drop model

1. The shape of the stable nucleus is spherical, the liquid drop is also spherical due to surface tension.
2. The intermolecular force of attraction in the liquid drop is identical to the nuclear force within the nucleus because both are saturated, short ranged etc.
3. The liquid molecules will move randomly within the liquid drop. Similarly, nucleons will move randomly within the nucleus.
4. The binding energy of the nucleus is constant. Similarly, latent heat of vaporization of a liquid drop is also constant.
5. The process of absorption of the foreign particle by the nucleus is identical to the condensation of the liquid drop.
6. When the temperature of liquid drop is increased, surface molecules will evaporate due to thermal agitation of molecules. Similarly, when energy is given to the nucleus through bombarding particles,it can emit the particle like alpha,beta,gamma etc immediately.
7. when a liquid drop is allowed to oscillate, it split into two or lighter drops which are identical to fission of the heavy nucleus into two or more lighter nucleus for better stability.

semi-empirical mass formula (weiz-sacker formula)

The formula connecting the mass of the nucleus and it’s binding energy is called semi-empirical mass formula. The stability of nucleus depends on upon the binding energy, which is the energy equivalent of the lost mass.

The binding energy (B) of a nucleus of mass number (A) and atomic no (Z) is defined by

B=(Zm_p + (A-Z)m_n –M)………..(1) ,where m_p and m_n are masses of proton and neutron

The binding energy is made by the following energy:

a) volume energy

we know that binding energy of any nucleus is directly proportional to it’s mass number.

i.e. B.E .α A

also, the volume of any nucleus is

volume α A.

therefore, volume energy (E_v) α A

so (E_v)=a₁ A………(2) where, a₁ is the proportionality constant whose value is 14 MeV

b)coulomb energy

This is due to electrostatic repulsion between the protons in the nucleus . As proton is the positively charged particle,any of the protons will interact with rest of the proton.

Consider a nucleus with Z protons and\(\frac{ Z(Z-1)}{2}\) proton- proton pairs.If ‘r’ be the separation between two protons,the coulomb energy

(E_C) between this two proton is found to be

(E_C)α \(\frac{ Z(Z-1)}{2}\)

therefore(E_C) = -a_{2\(\frac{ Z(Z-1)}{r}\)},, where a₂is proportionality constant whose value is 0.6 MeV and -ve sign indicates it tends to decrease the B.E. of nucleus.

c)surface energy

similar to the liquid drop, the nucleons inside the nucleus will be in random motion and uniformly distributed, but the interaction of surface nucleon will be less than that of inner nuclei.So an energy is created which is called surface energy and is found to be proportional to the surface area of the nucleus.

i.e. E_sα areaα r²

E_s= -a₃ A^2/3 [since rα A^1/3] ,, wherea₃isis proportionality constant whose value is 13 MeV. and -ve sign indicates that surface distribution is in opposite to volume energy.

1d) asymmetry energy

The nucleus having an equal number of protons and neutrons is found to be stable, which is called surface energy and neutron is found to be stable which is called symmetry effect. Any deviation from these symmetry makes the nucleus unstable. The energy due to asymmetry (B_a) is found to be

asymmetry (B_a) α \(\frac{(N-Z)^2}{A}\) =\(\frac{(a-2Z)^2}{A}\)

therefore (B_a) = \(\frac{-a_a(a-2Z)^2}{A}\) ,where a_ais constant having value 19 Mev.

1. e) Pairing energy

Nucleons have the tendency to exist pair ,Hence the nucleon with even no.z of protons even noumber N of neutrons have highly stable ,even z-odd. N of neutrons or odd z-even n are less stable . The nucleons in a nucleus exist in the form of the shell which may be filled by a certain number given by selection rules .Taking both effects together ,the term contributing towards volume energy is

$$B_p= a_pA^{-3/4}\;\;\;\;\;\; \dots \dots (i)$$
$$\;\; where\; a_p= 34 MeV \;for\; even \;A-even Z$$

$$\;\; =-34 MeV\; for\; even\; A-odd Z$$

$$\;\;\;= 0 \;for \;odd \;A$$

Thus total binding energy is

$$B=a_vA- a_sA^{2/3}-a_c\frac{z^2}{A^{1/3}}- a_a\frac {(A-2Z)^2}{A} +a_pA^{-3/4}\;\;\;\;\;\ \dots \dots(ii)$$

$$\frac{B}{A} =a_v- \frac{a_s}{A^{1/3}}-a_c\frac{z^2}{A^{4/3}}- a_a\frac {(A-2Z)^2}{A^2} +a_p\frac{A^{-3/4}}{A}$$

combining (i) and (ii) we get,

$$M(z,A) = Zm_p+(A-z)m_n-B$$

$$M(z,A) = Zm_p+(A-z)m_n-a_vA- a_sA^{2/3}-a_c\frac{z^2}{A^{1/3}}- a_a\frac {(A-2Z)^2}{A} +a_pA^{-3/4}\;\;\;\;\;\ \dots \dots(iii)$$

Finally in terms of mass

$$M(z,A) = Zm_p+(A-z)m_n+\frac{1}{c^2}\left[-a_vA- a_sA^{2/3}-a_c\frac{z^2}{A^{1/3}}- a_a\frac {(A-2Z)^2}{A} +a_pA^{-3/4} \right]\;\;\;\;\;\ \dots \dots(iii)$$

Which is semi-empirical mass formula.