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 πR³

    ⁼ 4/3 π (R₀A^1/3)³

  = 4/3 πR₀³A

Here we have used Eq. R=R₀A^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, is

ρ_n = A / V 

    = [A] / [(4/3)(πR₀³A)]

   = 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 then

ρ = (0.138 nucleon/fm³) (1.67×10^-27 kg/nucleon) × (10¹⁵ fm/m)³

  ≈ 2×10¹⁷ kg/m³

Thus from the above observation, we conclude that, its density would be 2×10¹⁷ kg/m³ . This is about 2×10¹⁴ times the density of water.

Question 2:-

(a) How much energy is required to separate the typical middle-mass nucleus ¹²⁰Sn into its constituent nucleons?

(b) What is the binding energy per nucleon for this nuclide?

Solution:-

(a) We can find this energy from Q = Δmc². 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  ¹²⁰Sn (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 is 

m = (50×1.007825 u) + (70×1.008665 u)

   = 120.997 80 u

This exceeds the atomic mass of ¹²⁰Sn by 

Δm = 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 ¹²⁰Sn nucleus is separated into 50 (bare) protons and 70 neutrons. In energy terms this mass difference becomes

Q = Δmc²

   = (1.096 u) (931.5 MeV/u)

   = 1021 MeV

Therefore, the energy is required to separate the typical middle-mass nucleus ¹²⁰Sn 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 then

E_n = Q/A = 1021 MeV/120 = 8.51 MeV/nucleon

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

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 and m_s represent the nuclear masses ³²P and ³²S and let the italic symbols m_p and m_s represent their atomic masses. We take the disintegration energy Q to be Δmc², where, from Eq. ³²P→³²S + e^- + ν  ( = 14.3 d),

Δm = 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 obtain

 Δm = (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  ³²P decay is then 

Q = Δmc²

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

   = 1.71 MeV

Therefore, from the above observation we conclude that, the disintegration energy Q for the beta decay of ³²P would be 1.71 MeV.