Find the Binding energy per nucleon for $$^{120}_{50}Sn$$. Mass of proton $$m_p = 1.00783$$ U, mass of neutron $$m_n = 1.00867$$ U and mass of tin nucleus $$m_{sn} = 119.902199$$ U. (take 1U = 931 MeV)

For the nucleus $$^{120}_{50}\text{Sn}$$ we first note the numbers of each kind of nucleon. We have $$Z = 50$$ protons and $$N = 120-50 = 70$$ neutrons.

To obtain the binding energy we need the mass defect. The mass of all the free (unbound) nucleons is

$$m_\text{free} = Z\,m_p + N\,m_n.$$

Substituting the given numerical values,

$$m_\text{free} = 50\,(1.00783\;\text{u}) + 70\,(1.00867\;\text{u}).$$

Evaluating each term separately,

$$50 \times 1.00783 = 50.3915\;\text{u},$$

$$70 \times 1.00867 = 70.6070\;\text{u}.$$

Adding these,

$$m_\text{free} = 50.3915\;\text{u} + 70.6070\;\text{u} = 120.9985\;\text{u}.$$

The actual measured mass of the tin nucleus is given as

$$m_\text{nucleus} = 119.902199\;\text{u}.$$

Hence the mass defect is

$$\Delta m = m_\text{free} - m_\text{nucleus} = 120.9985\;\text{u} - 119.902199\;\text{u} = 1.096301\;\text{u}.$$

The binding energy formula is stated as

$$E_B = \Delta m\,c^2,$$

and using the conversion $$1\;\text{u} = 931\;\text{MeV}/c^2,$$ we write

$$E_B = 1.096301\;\text{u} \times 931\;\text{MeV/u}.$$

Multiplying,

$$E_B = 1.096301 \times 931 = 1020.656231\;\text{MeV}.$$

To find the binding energy per nucleon, we divide by the mass number $$A = 120$$:

$$\frac{E_B}{A} = \frac{1020.656231\;\text{MeV}}{120} \approx 8.505\;\text{MeV}.$$

Rounded to one decimal place, this is $$\approx 8.5\;\text{MeV}.$$/p>

Hence, the correct answer is Option D.