Quantum Physics and Nuclear Physics

Determining the Mass of an Atom Nuclear Physics Determining the Mass of an Atom If a charged particle moves at a constant speed through a magnetic field it will experience a force (=Bqv) that provides a centripetal force (=mv2/r)to make it move through circular motion: Thus Clearly if velocity, charge and flux density are constant then the mass of the particle is proportional to the radius of the circle it moves through. Bqv = mv2 r m = Bqr v 1

The Bainbridge Mass Spectrometer In a Bainbridge mass spectrometer, positive ions are created and injected into the velocity selector. Here the positive ions are subjected to both magnetic and electric fields. These create forces that act in opposite directions.

Photographic plate

If the two forces are balanced the particle will travel in a straight line through the exit slit (s2 in the diagram)... Bqv = Eq so the only particles to exit the velocity selector are those with velocity equal to the ratio of the Electric field strength (E) to the magnetic flux density (B)... Sub into equation 1 gives: v = E B m = B2qr E

Q1. The electric field strength between the plates of the velocity selector in a Bainbridge mass spectrometer is 1.20 x 105 Vm-1, and the magnetic field in both regions is 0.600 T. A stream of singly charged neon atoms moves in a circular path of 0.728 m radius in the magnetic field. Determine the mass of one neon atom the mass number of this neon isotope. (1 amu = 1.66 x 10-27kg) a. 3.50 x 10-25 kg; b. 21

Q2. In the Bainbridge mass spectrometer, suppose the magnetic field B in the velocity selector is 1.30 T, and ions having a speed of 4.00 x 106 ms-1 pass through undeflected. What is the electric field between the plates? If the separation of the plates is 0.50 cm, what is the potential difference across the plates? If an ion of mass 5.30 x 10-26 kg passes into the deflection chamber, calculate its radius of deflection (assume q = 1.60 x 10-19C). a) 5.20 x 106 V/m; b) 2.60 x 104 V; c) 1.02m

Determining the Size of a Nucleus When a positive charged particle approaches a nucleus such as in scattering experiments (like Rutherford’s alpha particle scattering experiment) it is repelled by electrostatic forces. The distance of closest approach (d) is taken as being a measure of the radius of the nucleus (it has the same order of magnitude). Nucleus α d α v

Assuming all the initial KE of the alpha particle has been converted into electric potential energy: so... As all particles position is uncertain, this gives only a rough measure of the nuclear size. ½ mv2 = kQq r Q = charge on nucleus = Ze q = charge on alpha = 2e k = constant ( Z = proton number e = charge on electron) r = k Ze 2e ½ mv2 r = 4kZe2 mv2 Note: the Fermi model says that r = r0A1/3 where r0 = 1.2 x 10-15m = 1.2 fm (Note that the femtometre shares the unit symbol (fm) with the older non-SI unit fermi, to which it is equivalent. The fermi, named in honour of Enrico Fermi, is often encountered in nuclear physics.)

Nuclear Energy Levels It is not just the electrons that have energy in an atom. The nucleus does too. When alpha particles are emitted by a nucleus they are observed to have discreet energies. Also, gamma photons are emitted with discreet spectra. These observations suggests that the nucleus must have energy levels, similar to atomic energy levels. We say a nucleus that is capable of emitting a gamma (or alpha or beta) particle is unstable. This means it is in an excited state, elevated to a higher nuclear energy level. It can become more stable by emitting a gamma particle and thus falling to an energy level closer to its ground state.

This image shows how possible changes in the excitation of a Nickel nucleus could lead to it emitting gamma photons of discreet frequencies. Note the difference in the values of energy compared to atomic energy levels. Q. Explain the differences in the emission of X rays and gamma rays from the atom.

Beta Emission Beta particles are not emitted with discreet energies. They have a continuous range of energy values:

Thus the Beta particle can have a continuous range of energies. To account for this in 1930 Wolfgang Pauli postulated the existence of the neutrino ( - nu), not detected until 1956. When the nucleus falls to a lower energy level the energy emitted is shared between the Beta particle and the neutrino. Thus the Beta particle can have a continuous range of energies. Pauli originally named his proposed light particle a neutron. When James Chadwick discovered a much more massive nuclear particle in 1932 and also named it a neutron, this left the two particles with the same name. Enrico Fermi, who developed the theory of beta decay, coined the term neutrino in 1934 as a clever way to resolve the confusion. It was a pun on neutrone, the Italian equivalent of neutron: neutrone seems to use the -one suffix (even though it is a complete word, not a compound), which in Italian indicates a large object, whereas -ino indicates a small one.

Na → Ne + e +  Beta plus decay A beta plus particle is a positive electron (a positron, a particle of anti-matter). When a proton turns into a neutron in the nucleus the positron is emitted with a neutrino (). E.g. 22 11 22 10 Na   →   Ne  +  e +    +1

Exponential Decay Decay of a radioactive sample is an exponential process. This means the rate at which the nuclei decay (the activity (A) of the sample) is proportional to the number of nuclei remaining at any instant. We can also say... Activity = -  N = decay constant (s-1) N = No. of radioactive nuclei at a certain time No = No. of radioactive nuclei at beginning (t=0) dN = -  N dt N = N0 e– λt A = A0 e– λt

Probability of Decay The bigger the decay constant the quicker a substance decays. If N = 1 then dN = -  N dt dN = -  dt - the decay constant gives us a measure of the probability of any single nucleus decaying in unit time.

Decay constant and half life After t = t½ there will be half the original number of radioactive nuclei. i.e. N = ½ N0 so ... taking logs... this the same as... ½ N0 = N0 e- λt½ ½ = e- λt½ ln ½ = - λt½ ln 2 = λt½ t½ = ln 2 λ

Determination of Half Life Experiment: Using a simulation to determine half life Method: - use a fresh sample of protactinium (a radioactive nuclide with a short half life) determine values of activity every ten seconds. plot a suitable graph to determine the decay constant and thus calculate half life.

Note: to plot a straight line graph take logs of both sides of the activity equation: A = A0 e– λt ln A = ln A0 - λt ln A = - λt + ln A0 ( y = mx + c ) lnA t m = - λ

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