# Nuclear Physics Test

1. Compute the average energy per nucleon for the $He$ nucleus. Numerical data: $M_p =1.00782522 amu$, $M_n =1.00866522 amu$, $M_{He} = 4.0026033 amu$.

$^4_2He\rightarrow 2*^1_1p +2*^1_0n$

$B =(2*1.00782522 +2*1.00866522 -4.0026033)/4 =0.007595 amu$

$1 amu = 931.5 MeV$

$B = 7.074 MeV$

2. Compute the separation energy of the $\alpha$ particle from the $^{16}O$ nucleus. Numerical data: $M_O =15.9949150 amu$, $M_{He} =4.0026033 amu$, $M_C =12 amu$.

$^{16}_8O\rightarrow ^{12}_6C +^4_2\alpha$

$W =12+4.0026033 -15.9949150 =0.00769 amu =7.1617 MeV$

($1 amu =931.5 MeV$)

3. What conclusion results from the affirmation: “nucleons keep their original volume when they are bind together in a nucleus”?

The conclusion is that the density of nucleus remains constant, not depending on its size.

4. What properties of nuclear forces result from the affirmation: “nuclei are stable and have almost a spherical shape”?

Because the nuclei are stable this means that nuclear forces are attractive and more intense than electric forces. Because the nuclei are spherical it results that nuclear forces have a “saturation” character, that is they act on short distances (like the interaction forces between water molecules in a spherical drop).

5. What arguments can you raise in the favor of the “liquid drop” model of a nucleus?

a) The constant density of nucleus means that basically the “nuclear matter” is incompressible (like liquids).

b) The saturation character of nuclear forces is like the intermolecular forces in a liquid (see answer to question 4).

c) The value of the bonding energy per nucleon is about the same for all nuclei, which means all nucleons interact one which each other with about the same force that do not depend on the nucleus size or on the type of nucleon (proton or neutron). (like the molecules from a liquid)