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**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}3A

Here 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 ρn, expressed in nucleons per unit volume, is

ρ_{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 then

ρ = (0.138 nucleon/fm^{3}) (1.67×10^{-27} kg/nucleon) × (10^{15} fm/m)^{3}

≈ 2×1017 kg/m^{3}

Thus from the above observation, we conclude that, its density would be 2×10^{17} 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?**

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

m = (50×1.007825 u) + (70×1.008665 u)

= 120.997 80 u

This exceeds the atomic mass of

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

Q = Δmc

= (1.096 u) (931.5 MeV/u)

= 1021 MeV

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

(b) The total binding energy Q is the total energy that must be supplied to dismantle the nucleus. The binding energy per nucleon E

E

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 ^{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.**

Because of the presence of the emitted electron, we must be careful to distinguish between nuclear and atomic masses. Let the boldface symbols mp and ms represent the nuclear masses

Δm = m

in which me is the mass of the electron. If we add and subtract 15me on the right side of this equation, we obtain

Δm = (m

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

Q = Δmc

= (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

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