Solved Examples on Nuclear Physics

Question 1:-We can think of all nuclides as made up of a neutron-proton mixture that we can call nuclear matter. What is its density?

Solution:-We know that this density is high because virtually all the mass of the atom is found in its tiny central nucleus. The volume of the nucleus (assumed spherical) of mass number A and radius R is

V = 4/3 πR

^{3}

^{ = }4/3 π (R_{0}A^{1/3})^{3}= 4/3 πR

_{0}^{3}AHere we have used Eq. R=R

_{0}A^{1/3 }to obtain the third expression. Such a nucleus contains A nucleons so that its nucleon number density ρ_{n, }expressed in nucleons per unit volume, isρ

_{n }= A / V= [A] / [(4/3)(πR

_{0}^{3}A)]= 3/[(4π) (1.2 fm)

^{3}]= 0.138 nucleon/fm

^{3.}We can consider nuclear matter as having a single density for all nuclides only because A cancels in the equation above.

The mass of a nucleon (neutron or proton) is about 1.67 x 10

^{-27}kg. The mass density of nuclear matter in SI units is thenρ = (0.138 nucleon/fm

^{3}) (1.67×10^{-27}kg/nucleon) × (10^{15}fm/m)^{3}≈ 2×10

^{1}^{7}kg/m^{3}Thus from the above observation, we conclude that, its density would be 2×10

^{1}^{7}kg/m^{3 }. This is about 2×10^{14 }times the density of water.

Question 2:-(a) How much energy is required to separate the typical middle-mass nucleus

^{120}Sn into its constituent nucleons?(b) What is the binding energy per nucleon for this nuclide?

Solution:-(a) We can find this energy from Q = Δmc

^{2}. Following standard particle, we carry out such calculations in terms of the masses of the neutral atoms involved, not those of the bare nuclei. One atom of^{120}Sn (nucleus plus 50 electrons) has a mass of 119.902 199 u. This atom can be separated into 50 hydrogen atoms (50 protons, each with one of the 50 electrons) and 70 neutrons. Each hydrogen atom has a mass of 1.007825 u, and each neutron a mass of 1.008665 u. The combined mass of the constituent particle ism = (50×1.007825 u) + (70×1.008665 u)

= 120.997 80 u

This exceeds the atomic mass of

^{120}Sn byΔm = 120.99780 u – 119.902 199 u

= 1.095601 u

≈ 1.096 u

Note that since the masses of the 50 electrons cancel in the subtraction, this is also the mass difference that is obtained when a bare

^{120}Sn nucleus is separated into 50 (bare) protons and 70 neutrons. In energy terms this mass difference becomesQ = Δmc

^{2}= (1.096 u) (931.5 MeV/u)

= 1021 MeV

Therefore, the energy is required to separate the typical middle-mass nucleus

^{120}Sn into its constituent nucleons would be 1021 MeV.(b) The total binding energy Q is the total energy that must be supplied to dismantle the nucleus. The binding energy per nucleon E

_{n}is thenE

_{n}= Q/A = 1021 MeV/120 = 8.51 MeV/nucleonFrom the above observation we conclude that, the binding energy per nucleon for this nuclide would be 8.51 MeV/nucleon.

Question 3:-Calculate the disintegration energy

Qfor the beta decay of^{32}P, as described by equation^{32}P→^{32}S + e^{-}+ ν ( = 14.3 d). The needed atomic masses are 31.97391 u for^{32}P and 31.97207 u for^{32}S.

Solution:-Because of the presence of the emitted electron, we must be careful to distinguish between nuclear and atomic masses. Let the boldface symbols

m_{p }andm_{s}represent the nuclear masses^{32}P and^{32}S and let the italic symbolsm_{p}andm_{s }represent their atomic masses. We take the disintegration energyQto be Δmc^{2}, where, from Eq.^{32}P→^{32}S + e^{-}+ ν ( = 14.3 d),Δ

m =m_{p}– (m_{s}+m_{e}),in which

m_{e}is the mass of the electron. If we add and subtract 15m_{e}on the right side of this equation, we obtainΔ

m =(m_{p}+ 15m_{e}) – (m_{s}+ 16m_{e})The quantities in parentheses are the atomic masses in this way, the mass of the emitted electron is automatically taken into account. (This will not work for positron emission.)

The disintegration energy for the

^{32}P decay is then

Q= Δmc^{2}

=(31.97391 u – 31.97207 u) (931.5 MeV/u)= 1.71 MeV

Therefore, from the above observation we conclude that, the disintegration energy

Qfor the beta decay of^{32}P would be 1.71 MeV.