The semi empirical mass formula is used to calculate the binding energy of the nucleus. It is given as:
$$B.E = a_vA - a_sA^{2/3} - a_c\frac{Z^2}{A^{1/3}} - a_a\frac{(A-2Z)^2}{A} + \delta(A,Z)$$
where,

• B.E = Binding energy of the nucleus


• a_v, a_s, a_c, a_a = Constants


• A = Mass number


• Z = Atomic number


• Δ(A,Z) = Correction factor

To find the most stable isobar for a nucleus having atomic number 97, we need to calculate the binding energy of all the isotopes with atomic number 97 and choose the one with the highest binding energy. The isotope with the highest binding energy will be the most stable isobar.


There are three isotopes with atomic number 97:

• Technetium-97 (A=97, Z=43)


• Rhodium-97 (A=97, Z=45)


• Palladium-97 (A=97, Z=46)

Using the semi empirical mass formula, we can calculate the binding energy of these isotopes. The constants for the formula are:

• a_v = 15.67 MeV


• a_s = 17.23 MeV


• a_c = 0.75 MeV


• a_a = 93.2 MeV

The correction factor Δ(A,Z) is given by:

• Δ(A,Z) = 0, if A-Z is odd


• Δ(A,Z) = (-1)^A-Z, if A-Z is even

Using these values, we can calculate the binding energy of each isotope:

Technetium-97:
$$B.E = (15.67\times97) - (17.23\times97^{2/3}) - (0.75\times43^2/97^{1/3}) - (93.2\times(97-2\times43)^2/97) + \delta(97,43)$$
$$B.E = 1372.8\text{ MeV}$$

Rhodium-97:
$$B.E = (15.67\times97) - (17.23\times97^{2/3}) - (0.75\times45^2/97^{1/3}) - (93.2\times(97-2\times45)^2/97) + \delta(97,45)$$
$$B.E =