Are more massive nuclei wider? How does nuclear radius depend on mass? How dense is the nucleus?

Nuclear Radius In 9.1, by using probability and the Least Distance Approach method the diameter of the nucleus is in the order of 10-15m (1 fentometre). Estimated to be about 10fm for gold, 7fm for Aluminium. See p150. d = D/10 000 diameter d, approx 104 – 105 smaller than atom EK = QαQN 4πε0d d See table p89

Electron Diffraction intensity θmin angle of diffraction A more accurate method fires a beam of high energy electrons (through a pd of 100 million volts) at a thin sample. Electron wavelength depends on their speed. The electrons are diffracted by the nuclei because these electrons have a de Broglie wavelength similar to, but smaller than the diameter of the nucleus (10-15m). A detector shows that as the angle of the detector from the “zero order” beam (the normal) is increased, the intensity of the diffracted beam decreases. This is because the electrons are attracted by the nuclei. Diffraction by each nucleus causes intensity maxima and minima to be superimposed on to this pattern, similar to the fringes seen with light, (as long as the de Broglie wavelength is no greater than the diameter of the nucleus). The angle from the centre of the first minimum, θmin and the wavelength are used to calculate the diameter, (see p 176). The intensity distribution in the diffraction pattern for a large number of slits is shown

R = r0A1/3 where A = mass number and constant r0 = 1.05fm. By sampling different elements, it can be shown that the radius R is proportional to the cube root of the mass A R = r0A1/3 where A = mass number and constant r0 = 1.05fm. (see data sheet) So yes, more massive nuclei are wider 64Cu 32S 16O The radius of copper is 21/3 (1.26) x bigger than sulphur and 41/3 (1.59) x bigger than that of oxygen

R = r0A1/3 If you were given values of R and A for a range of elements, what straight line graph(s) could you plot to confirm the above relationship and how would you find the value of r0? See p 177 for graph solutions and Exam Q 9 on p181.

Plot R against A1/3, goes through origin with gradient r0 Plot lnR against lnA to confirm the power (gradient) and y intercept lnR = lnA1/3+ lnr0 lnR = 1/3 lnA + lnr0 By cubing… R3 = r03A Plot R3 against A to get gradient r03

Nuclear Density Assuming the nucleus is spherical, V = 4/3πR3 = 4/3π(r0A1/3)3 Hence V = 4/3πr03A (density = A/V) So: the nuclear volume is proportional to the mass A so the density must be constant. the density is constant throughout the nucleus and independent of the radius the nucleons are evenly and equally separated, regardless of the size of the nucleus.

1mm3 of nuclear matter = 340 million kg (the same mass of 4 million adults) A neutron star is almost as dense. Now try the Summary Qs p177 and Exam Style Qs p178-181

E = mc2 where m = change of mass Ch 10: Energy & Mass E = mc2 where m = change of mass The mass of an object increases when it gains energy. This applies to all energy changes, of any object, but is only really significant in nuclear reactions.

The reason why mass changes when energy is transferred is still not clearly understood, but… When a particle and antiparticle meet they annihilate each other and 2 gamma photons are produced, each with energy mc2. In pair production, a single gamma photon with energy > 2mc2 can produce a particle and an antiparticle, each of mass, m

Energy released Q = Δmc2 where no other energy is supplied. If we know the Rest Mass of each particle in a nuclear reaction, and their mass after: Energy released Q = Δmc2 where no other energy is supplied. Where energy is released, the total energy after must be less than before. 1 atomic mass unit, 1u = 1.661 x 10-27kg, so the energy change for 1u, E = 1.66 x 10-27 x (3.0 x 108)2 = 1.49 x 10-10J = 931.3MeV

1. Alpha decay In alpha decay, the nucleus recoils when α is released, so energy is shared in inverse proportion to their masses, (by conservation of momentum).

2. Beta decay In beta decay, the energy released is shared in variable proportions by the three particles. When the β particle has max kinetic energy, the neutrino or antineutrino has negligible Ek and its max Ek is only slightly less than the total released as the nucleus has such a large mass. When calculating Q in beta decay, assume the mass of the neutrino is negligible.

3. Electron capture The nucleus emits a neutrino which carries away the energy released in the decay The atom also emits an x-ray photon when the inner shell vacancy due to the electron capture is filled

The Strong Nuclear Force