Mass Parabola

The semi-empiricalmass formula helps to account for some of the important features of stabilityof nucleus ,in particular, $beta$-activity and stabilityproperties of isobar,

According to semi-empiricalmass formula ,mass of given nucleus (A,z) is

$$M(z,A) = Zm_p+(A-z)m_n-a_vA- a_sA^{2/3}-a_c\frac{z(z-1)}{A^{1/3}}- a_a\frac {(A-2Z)^2}{A} \pm B_p$$

$$=A\left[m_n-a_v +\frac{a_s}{A^{1/3}}+a_a \right]+\left(m_p-m_n-\frac{a}{A^{1/3}}-4a_a\right)z+\left(\frac{a_c}{A^{1/3}}+\frac{4a_a}{A}Z^2 \right)\pm B_p$$

$$\therefore M(A,Z) =\alpha A+\beta Z+\gamma Z^2 + \delta \;\;\;\;\;\; \dots \dots (i)$$

$$where$$

$$\alpha =[m_n-a_v +\frac{a_s}{A^{1/3}}+a_a]$$

$$\beta =(m_p-m_n-\frac{a}{A^{1/3}}-4a_a)$$

$$\gamma =(\frac{a_c}{A^{1/3}}+\frac{4a_a}{A})$$

And B_p(pairing energy )= +$\delta$ for even z and even N

= -$\delta$ for add z and odd N

= o for even z, odd N or vice - versa

This equation (i) is the equation of parabola .

i) for odd A nuclei:

For odd A, $\delta$ =0 so there is only one parabola implying that there is only one stable nuclei .The mass 'm' vs 'z' is ploted

for a series of nuclidies with constant odd A=91 and varying z as shown in figure .

From the figure, we see that stable nucleus is $_{40}Zr^{91}$.Isobar to a left of the stablenucleus have fewer protons{i.e larger mass )than the stable one and decay by $\overrline{\beta}$\) emission ie.

$$_{36}Kr^{91} \overset{\beta}{\rightarrow} _{37}Rb^{91}\overset{\beta }{\rightarrow} _{38}Sr^{91}\overset{\beta }{\rightarrow}_{39}Y^{91}\overset{\beta }{\rightarrow} _{40}Zr^{91}$$

Simillarly isobars to the right of stable isobar have excess proton that the stable one and decay by $\beta$⁺(positron) emission.(some time k capture)

$$_{42}Ma^{91} \overset{\beta }{\rightarrow} _{41}Nb^{91}\overset{\beta }{\rightarrow} _{40}Zr^{91}$$

from (i) we get

$$\left(\frac{\delta M}{\delta z}\right)_{A= constant\; odd}=\beta +2\gamma z$$

for stable isobar let z= z₀ ,then

$$\beta +2\gamma z =0$$

$$\therefore Z_0=\frac{-\beta}{2\gamma}\;\;\;\;\dots \dots (ii)$$

Putting (ii) in (i) we get

$$M= \alpha A + Z_0(-2 \gamma Z_0) +\gamma Z_0^2$$

$$= \alpha A-2 \gamma Z_0^2+\gamma Z_0^2$$

$$\therefore M(Z_0, A)=\alpha A-\gamma Z_0^2$$

Which gives the mass of the stable isobar .

Now again energy is released during $\beta$ emission is

$$Q_{\beta }^- = M(Z, A) - M(Z+1, A)$$

$$=[ \alpha A + \beta Z + \gamma Z^2]-[ \alpha A + \beta (Z+1)+ \gamma (Z+1)^2]$$

$$=-\beta -2\gamma Z-\gamma$$

$$=2\gamma Z_0 -2\gamma Z-\gamma\;\;\;\;\;\;\;\;\; (from \;(ii))$$

$$\therefore Q_{\beta }^- =2\gamma (Z_0-Z-\frac{1}{2})$$

Similarlly energy released in $\beta$^- decay is

$$Q_{\beta }^- = M(Z, A) - M(Z+1, A)$$

$$Q_{\beta }^- =2\gamma (Z_0-Z-\frac{1}{2})$$

(ii) For even A

For even A $\delta$≠ 0 ,so that we get two parabola didplsced in B.E.by 2 $\delta$. There can be three stable nuclei . The parabolic relation between the masses (Z,A) of the members of the isobari family is shown as in the figure .Energeticlly posible decays are as shown in figure.Notice 44 Ru can't $\beta$ - decay to 45 Rh as the masof 45 Rh is greater than the mass of 44Ru .We see that $P_d^{104}$ is stable having least mass in the family .Mass of Ru is greater than the mass of Pd .But this decay is again forbidden as mass of the internediate nucleus Rh is greater than the masss of Ru . Thus there are two stable nuclei 44 Pd¹⁰⁴

If Ru decays into Rh in a time .$$\Delta t= \frac {h}{\Delta E} (-10^{-22}sec ) ,then the violation of conservation of energy by an amount \(\Delta\)E is consistent with Heisenberg's uncertainty principle and provided Rh decays to pd with the time i.e. a double \(\beta\) -decay leading Ru to Pd is positive .Normally ,the time required for the normal(\beta\) -decay is many seconds .So that the probability for the decay taking place in \(-10^{-22}sec\) is extremely small and not yet observed experimentally .This is the reason why for even A nuclei there can be one ,two or three stable nuclei .

$$for \; A= 134 \rightarrow 3\; stable\; nuclei$$

$$=200 \rightarrow 1\; stable\;nuclei$$

Significance of mass-parabola

As shown in the figure the odd N-odd z nuclei lie on the upperand are therefore unstable with the respect to those on a lower curve :with the result that no stable odd-odd nuclei should exist .

The only exception to this rule are extremely light nuclei for eg; an isobaric family with A= 14 in this family only $_7N^{14}$ is the stable member the fact that light nuclei should not be treated on the basis of liquid drop model as they have too few nucleons to be treated as a liquid drop.