Symmetric Fission From liquid Drop Model Approach

We can write the fission reaction is

$$_ZX^A \longrightarrow _{Z_1}X_1^A + _{Z_2}X_2^A +Q $$

In symmetric fission reaction the nucleus \(_zX^A\)splits into two identucal nuclei \(_{Z/2}X_2^{A/2}\)

$$(Z_1=Z_2=\frac{Z}2 \;for \;symmetric\; )$$

i.e

$$_ZX^A \longrightarrow _{Z/2}X_1^{A/2} + _{Z/2}X_2^{A /2}+Q $$

Where Q is energy released in symmetric fission an is given by

$$Q= M(Z,A) - 2M \left(\frac{Z}2 \frac{A}2\right) \;\; \;\; \;\; \dots \dots (i) $$

And from the relation

$$(M,Z) = Z (m_p -m_n) +Am_n -B(Z,A) $$

$$and$$

$$ 2M \left(\frac{Z}2 \frac{A}2\right) =2. \frac{Z}2 (m_p-m_n)+ 2 \frac{A}2 m_n -2B\left( \frac{Z}2, \frac{A}2\right)$$

$$\therefore Q = 2B\left( \frac{Z}2, \frac{A}2\right)-B(Z,A)\;\;\;\;\;\;\;\; \dots \dots (ii) $$

Also from semiempirical mass formula

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

and$$2B\left( \frac{Z}2, \frac{A}2\right) =a_vA -a_s\frac{A^{2/3}}{2} -a_c \frac{Z/2(Z/2-1)}{{A/2}^{1/3}} a_a \frac{(A/2-2Z)^2}{A/2} +a_p \frac{A^{-3/4}}{2}$$

$$ \therefore Q=a_sA^{2/3}[1- 2\left(\frac{1}{2}\right)^{2/3}]-a_c \frac{Z(Z-1)}{A^{1/3}}\left(1-\frac{1}{2^{2/3}}\right) +pairing \; energy \; difference $$

The pairing energy difference is very small and can be neglected as an approximation.

$$\therefore Q= A^{2/3} \left[-0.26\; a_s +0.37\; a_c\frac{Z(Z-1)}{A}\right] \;\;\;\;\;\;\;\;\dots \dots (iv)$$

Thus Q to be greater than zero,

$$ A^{2/3} \left[-0.26\; a_s +0.37\; a_c\frac{Z(Z-1)}{A}\right]≥ \;0$$

$$\therefore \;\;\;\;\; \left(\frac{Z^2}{A}\right)≥\frac{0.26\;a_s}{0.37\; a_c}\;∼\;15\;\;\;\;\;\;\; (v) \;\;\;\;\;\;\ (\because a_s∼13 MeV ,a_c∼0.6 MeV)$$

Thus for all nuclei having Z> 35 and A>80 .Fission is possible and would release energy.

Equation (v) represented the required condition for symmetric fission. For the symmetric fission of \(_{92}U{238}\), using equation (iv) we get

$$\therefore Q= (238)^{2/3} \left[-3.38\; +0.22\frac{92(92-1)}{238}\right]-170MeV$$

Here the first term is the surface term which gives∼ -130 MeV and the second term is coulomb term which gives∼300 MeV. Here Q is +ve because in the large decrease in coulomb energy and increase in surface energy due to fission .The coulomb energy and surface energy together control the energetic of nuclear reaction.

Limit against symmetric spontaneous fission

We know the spontaneous fission is opposed by coulomb potential as shown in the figure .

The above plot shows how the mutual P.E of two fission fragments change when they approach each other .for r>2R the mutual P.E is simply coulomb and r=2R, the nuclear attractive force becomes active. At r= 2R the coulomb energy is

$$ E_c = \frac{1}{4\pi \epsilon _0} \frac{e^2(Z/2)^2}{2R}$$

$$ = \frac{1}{4\pi \epsilon _0} \frac{e^2(Z/2)^2}{2R_0(A/2)^{1/3}}$$

$$=0.262 a_c \frac{Z^2}{A^{1/3}}\;\;;\;\;\;\; \dots \dots (i)$$

$$a_c= \frac 35\frac{1}{4\pi \epsilon _0} \frac{e^2}{R_0}$$

For spontaneous fission

$$Q≥E_c$$

$$ie A^{2/3} \left[-0.26 a_s + 0.37 a_c \frac{Z^2}{A} \right]≥ 0.262a_c \frac{Z^2}{A^[1/3}$$

$$or, -0.26a_sA^{2/3} +0.37a_c \frac{Z^2}{A^{1/3}} ≥ 0.26 a_c \frac{Z^2}{A^[1/3}$$

$$or, 0.26a_s A+0.37a_c Z^2≥ 0.626 a_cZ^{2}$$

$$or,0.108 a_cZ^2≥ 0.26 a_sA$$

$$or, \frac{Z^2}A≥ \frac{ 0.26 a_s}{0.108 a_c}= 52 $$

This gives the stablity limit against symmetric, spontaneous fission. Here w have obtained the limit in which is too high as we find in the naturre elements having \(\frac{Z^2}A\) values around 44 or 50 are unstable against spontaneous fission. This is primarlly because it is not accurately known hoe the potential behaves around r=2R and less.

A much better estimate of the central elements of the critical value \(\frac{Z^2}{A}\)can be obtained as shown by Bohr and Wheeler by considering oscillation of a liquid drop under the joint action of surface force andcolumb forces.The figure shows how osvillation of this type will result into a division of a drop.

A drop is considered incompressible, the volume of the sphere of radius R is same as the volume of the ellipsoid of semi-major axis 'a' and semi-minor axis 'b' i.e.

$$\frac 43 \pi R^3 =\frac 43\pi ab^2 $$

if \(\epsilon\) denotes the eccentricity (\(\epsilon=0\;for\;sphere\))

$4a= R(1+ \epsilon ) and b= R(1 + \epsilon)^{-1/2}$$

Then the surface energy of ellipsoid is

$$B_s= Area\; of \; ellipsoid\; \times \; surface \; tension\;$$

$$= 4 \pi R^2 (1 + \frac 25 \epsilon ^2+ \dots \dots ) \times \; surface \;tension $$

$$\therefore B_s = a_s A^{2/3} \left(1+\frac 25\epsilon ^2\right) \dots \dots (A)$$

The coulomb energy of ellipsoid is

$$B_c= \frac 35 \frac 1{4\pi \epsilon } \frac{e^2Z^2}{R} (1-\frac 15 \epsilon ^2 + \dots \dots \dots )$$

$$B_c= a_c \frac{Z^2 }{ A^{1/3}} (1- \frac 15 \epsilon ^2) \dots \dots (B)$$

Clearly in the ellipsoidal form the coulomb energy is less compared to spherical form as the change are comparatively farther apart on the other hand the ellipsoid form has a larger surface energy term compares to the spherical form because of large surface area .

From A and B we take only terms in \(\epsilon\)

$$a_c \frac{Z^2}{A^{1/3}(\frac 15 \epsilon ^2 )≥ a_s A^{2/3} (\frac 25\epsilon ^2)$$

$$or, \frac{Z^2}{A}≥\frac{2a_s}{a_c} =44$$

$$0r ,\frac{Z^2}{A}≥ 44$$

This gives a much more acccurate relativistic stab;ity limit against spontaneous fission .