Nuclear Physics

Mass defect: the difference between the sum of the masses of the separate nucleons and the actual mass of the nucleus:

$\Delta m=\left[Zm_p+(A-Z)m_n\right]-M_{\text{nucleus}}$

Binding energy per nucleon: the average energy needed to remove one nucleon from the nucleus, i.e. the total binding energy divided by the mass number:

$\text{B.E. per nucleon}=\frac{\Delta m\,c^2}{A}$

A larger value means a more stable nucleus.

1 amu $=1.66\times10^{-27}\ \text{kg}$. By Einstein's mass–energy relation $E=mc^2$ with $c=3.0\times10^{8}\ \text{m s}^{-1}$:

$E=1.66\times10^{-27}\times(3.0\times10^{8})^2=1.49\times10^{-10}\ \text{J}$

Convert to electron-volts using $1\ \text{eV}=1.6\times10^{-19}\ \text{J}$:

$E=\frac{1.49\times10^{-10}}{1.6\times10^{-19}}=9.3\times10^{8}\ \text{eV}\approx931\ \text{MeV}$

(Using more precise values, $c=2.998\times10^{8}$ and $1\ \text{amu}=1.6605\times10^{-27}$ kg, gives $931.5$ MeV.)

(c) fermions

Fermions are the elementary particles that have half-integer spin (1/2, 3/2, ...). Quarks are fermions, while bosons have integer spin and hadrons are composite particles.

The nuclear radius is $R=R_0A^{1/3}$, so the volume is proportional to $A$ (the mass number). The mass is also proportional to $A$, since each nucleon has nearly the same mass. Hence the ratio mass/volume is independent of $A$, giving a nearly constant density of about $2.3\times10^{17}\ \text{kg/m}^3$ for all nuclei.

Protons repel each other through the long-range electrostatic force, which acts between every pair of protons in the nucleus. The attractive strong nuclear force has a very short range and acts between all nucleons (p-p, n-n, p-n). Neutrons add strong attraction without adding any electrical repulsion, so extra neutrons dilute the repulsion and help hold the nucleus together. As the nucleus gets bigger the repulsion grows faster, so heavier stable nuclei need more neutrons than protons ($N/Z$ rises from 1 to about 1.5).

(a) One atomic mass unit (u) is $\tfrac1{12}$ of the mass of a carbon-12 atom, $1\ \text{u}=1.66\times10^{-27}\ \text{kg}$.

Using $E=mc^2$:

$E=1.66\times10^{-27}\times(3\times10^{8})^2=1.49\times10^{-10}\ \text{J}=\frac{1.49\times10^{-10}}{1.6\times10^{-13}}\ \text{MeV}\approx931\ \text{MeV}$

So $1\ \text{u}\equiv931\ \text{MeV}$.

(b) Number of atoms in 1 g of U-235:

$N=\frac{1}{235}\times6.0\times10^{23}=2.55\times10^{21}$

Energy released: $E=2.55\times10^{21}\times3.2\times10^{-11}=8.2\times10^{10}\ \text{J}$.

Spread over one day ($86400\ \text{s}$):

$P=\frac{8.2\times10^{10}}{86400}\approx9.5\times10^{5}\ \text{W}\ (\approx0.95\ \text{MW})$

(c)

(a) Mass number $A=87+133=220$, atomic number $Z=87$. Notation: $^{220}_{87}\text{Fr}$.

(b) Energy per fission: $E=200\ \text{MeV}=200\times10^{6}\times1.6\times10^{-19}=3.2\times10^{-11}$ J.

Power $1\ \text{kW}=1000$ J s$^{-1}$, so the number of fissions per second is

$n=\frac{P}{E}=\frac{1000}{3.2\times10^{-11}}\approx3.1\times10^{13}\ \text{s}^{-1}$

(c) Balance the mass number $A$ and charge number $Z$ on both sides.

(i) $^{4}_{2}\text{He}+{}^{7}_{3}\text{Li}\rightarrow{}^{1}_{0}\text{n}+{}^{10}_{5}\text{B}$ (here $A=4+7-1=10$, $Z=2+3-0=5$, boron).

(ii) $^{2}_{1}\text{H}+{}^{16}_{8}\text{O}\rightarrow{}^{4}_{2}\text{He}+{}^{14}_{7}\text{N}$ (here $A=2+16-4=14$, $Z=1+8-2=7$, nitrogen).

(a) Rutherford directed fast $\alpha$-particles at a thin gold foil and observed the scattering with a ZnS screen.

• Most particles passed straight through, so most of the atom is empty space. The nucleus occupies only a tiny fraction of the atom, so it is very small.
• A very few particles (about 1 in 8000) were deflected through angles above $90^\circ$, some almost $180^\circ$. A heavy, fast $\alpha$-particle can be turned back only by a very strong force from a concentrated mass, so the nucleus is very dense.
• The $\alpha$-particles are positive and were repelled, so the nucleus is positively charged.

(b) The question gives $2.00\times10^{10}\ \text{J}$, which is impossible. The intended energy is $K=2.00\times10^{-13}\ \text{J}$. At the distance of closest approach all the kinetic energy becomes electrostatic potential energy:

$K=\frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{r}\ \Rightarrow\ r=\frac{2Ze^{2}}{4\pi\varepsilon_0K}$

$r=\frac{(9\times10^{9})(2)(79)(1.60\times10^{-19})^{2}}{2.00\times10^{-13}}\approx1.8\times10^{-13}\ \text{m}$

(c) This is only the distance of closest approach. The $\alpha$-particle stops and turns back while still outside the nucleus, because the Coulomb repulsion is still acting. It never touches the nucleus, so the nuclear radius is smaller than this value. The actual radius is about $7\times10^{-15}\ \text{m}$.

Rutherford's α-scattering experiment: A beam of fast-moving α-particles (helium nuclei, charge +2e) was directed at a very thin gold foil, and the scattered particles were detected by a movable zinc-sulphide screen.

Observations and conclusions:

1. Most α-particles passed straight through the foil — this shows that most of the atom is empty space. So the atom is largely hollow, and the nucleus occupies only a very small fraction of the atomic volume. Hence the nucleus is very small (radius of the order of $10^{-15}$ m, about $10^5$ times smaller than the atom).

2. A few α-particles were deflected through large angles, and about 1 in 8000 bounced back (nearly 180°) — a heavy, fast α-particle can be turned back only by a strong repulsive force acting over a very short distance. This is possible only if the positive charge and almost the whole mass of the atom are concentrated in a tiny central core. Hence the nucleus is very dense (nuclear density $\approx 10^{17}\ \text{kg m}^{-3}$) and carries the entire positive charge.

3. The deflected α-particles are repelled, not attracted — since α-particles are positively charged, the repulsion proves that the central core is positively charged.

Thus the experiment established that the atom has a tiny, massive, positively charged nucleus surrounded by a large empty space in which the electrons revolve.

The energy given in the question ($2.00\times10^{10}\ \text{J}$) is impossible for an $\alpha$-particle. The intended value is $K=2.00\times10^{-13}\ \text{J}$.

At the distance of closest approach all the kinetic energy becomes electrostatic potential energy:

$K=\frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{r}\ \Rightarrow\ r=\frac{2Ze^{2}}{4\pi\varepsilon_0K}$

$r=\frac{(9\times10^{9})(2)(79)(1.60\times10^{-19})^{2}}{2.00\times10^{-13}}=1.82\times10^{-13}\ \text{m}$

The maximum size (radius) of the gold nucleus is about $1.8\times10^{-13}\ \text{m}$.

The value in 11(b) is the distance of closest approach. It is the point where the $\alpha$-particle is momentarily at rest, turned back by the Coulomb repulsion, while still outside the nucleus. It never touched the nucleus. The distance is also measured between centres, so the $\alpha$-particle's own size is included. The nuclear radius is therefore smaller than this value, and the real radius of the gold nucleus is about $7\times10^{-15}\ \text{m}$.