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Revision:Atomic and Nuclear Physics 2

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Physics > Atomic and Nuclear Physics

13.1 Atoms and their constituents

13.1.1

Millikan's oil drop experiment involved first producing small droplets of oil with an atomizer. Some of these then fell through a small hole, and into a region between two fields. Using a variable resistance, the strength of the field was adjusted until the upward force of the field (The +ve plate was up the top, -ve on the bottom, but could be reversed for drops of opposite charge) equals the downward force of gravity. after it was balanced, the voltage was recorded, and then the drop was allowed to fall. After falling some distance, the drop falls at a constant speed (where the force of gravity is equaled by the air resistance). This terminal speed is measured, and allows the mass to be found using Stoke's law.

Then, we equate F_g and F_f , as follows:

$\displaystyle qE = mg \Rightarrow q = \frac{mg}{E}$ .

Since the mass of the drop can be found, and both and are known, we can find the charge of the drop. By graphing these charge values, we find that the smallest difference between them is e, 1.6 x 10^-19C.

13.1.2

Thompson's experiment is based around the counteracting effects of electric and magnetic fields. First, the two fields are adjusted so the beam passes undeflected between them, with the deflection effects of each equaling out ... Therefore: F_e = F_b , and so Eq = Bqv therefore $v = \frac{E}{B}$ .

The electric field is then removed, meaning the beam is deflected downwards by the magnetic field. Since the force from a magnetic field = evB , from $F_c = \frac{mv^2}{r}$ we get $evB = \frac{mv^2}{r}$ . This simplifies down to $\frac{e}{m} = \frac{v}{B_r}$ . We can measure the radius of curvature, we know the velocity from above, and we know the strength of the field, and so the value of $\frac{e}{m}$ can be found.

13.1.3

The distance of closest approach of a particle to the nucleus can be found by the conservation of energy. The initial energy of the particle is defined by $E_k = \frac{1}{2}mv^2$ . This energy is converted into potential energy as the charge approaches the nucleus. The potential energy at a given point is equal to the work done to move the charge to that point, so $W = q_{\alpha}V$ .

for a radial field is $\frac{1}{4 \pi E_o} \times q_{\mathrm{nucleus}}{d}$ , so the potential energy is $q_{\alpha} q_{\mathrm{nucleus}}{4 \pi E_o d}$ .

When this is equated with the kinetic energy, all the terms are known except the distance, which can therefore be found.

13.2 Nuclei and their constituents

13.2.1 : The mass spectrometer

First we start with a source of ions, all with a +1 charge. These are accelerated through an electric field. These ions then enter a velocity selector, an area with both magnetic and electric fields applying a force in opposite directions, so as to cancel each other out. This, as above the two forces equal out, and we get F_e = F_b , and so Eq = Bqv therefore $v = \frac{E}{B}$ . Thus, only ions with a particular speed are allowed through to the next stage. The electric field ends, and the ions are deflected through a magnetic field, resulting in a circular path. This means there is a centripetal force supplied by the field, and so $\frac{mv^2}{r} = Bqv$ , which rearranges to $m = \frac{Bqr}{v}$ . Assuming the field is varied to keep the radius constant, mass is proportional to magnetic field strength, as everything else is constant.

13.2.2 : How Chadwick discovered the neutron

Chadwick discover the effect by bombardment of Beryllium with alpha particles. A wax block (with its chain of hydrocarbons it serves as a proton source) is placed after a short distance, and protons are detected after the wax block. Thus there must be some way to 'knock out' the proton. From the energy the protons are in, the Compton effect due to possible gamma ray emission is too low to compensate. If the particle that alpha knocked out is a neutron though, it fits perfectly both by the conservation of energy and the conservation of momentum (since the p and n have approx. same mass, we would expect the 'pool players' result') Yet these particles are indeed uncharged. Further more, when these are to collide with protons in cloud chambers, a right angle track would result.

13.2.3 : Beta decay and the neutrino

The mechanisms are:

$\displaystyle p \longrightarrow n + e^+ + \nu$

( $\nu$ is a neutrino) and

$n \longrightarrow p + e^- + \overline{\nu}$

( $\overline{\nu}$ is an antineutrino).

13.2.4 : Decay Chains

Everything can be written as a series of nuclear equations, such as:

$\displaystyle {}^A_ZX + {}^{ 0}_{-1}e \longrightarrow {}^A_{Z-1}Y$ ,

the totals must be kept constant on both sides, but it's fairly easy.

13.2.5

The equation:

$\displaystyle N = N_0 e^{-\lambda t}$

is in the data book, while:

$\displaystyle \frac{\Delta N}{\Delta t} = -\lambda N$

is not. Both can be used to find decay stuff. is the number of radioactive nuclei, No is the starting number and $\lambda = \frac{\ln 2}{\mathrm{half life}}$ (that's in the data book). $\frac{\Delta N}{\Delta t}$ is the rate of decay.

13.2.6

Deduce:

$\displaystyle \lambda = \frac{\ln 2}{\mathrm{half life}}$ .

We start by taking a time after one half life, therefore $N = \frac{N_0}{2}$ . Therefore, $N = N_0 e^{-\lambda t}$ becomes $2 = e^{-\lambda t}$ . We then ln both sides to get $\ln 2 = \lambda t_{0.5}$ , and this rearranges to the required expression.

13.2.7

By coulomb's law, the protons in the nucleus should all repel each other and break it to bits. Therefore, there must be another force holding it all together, called the strong nuclear force. This is a force which greatly outweighs the electromagnetic repulsion, but only acts over a very small distance (within the nucleus).

13.3 Energy changes within atoms

13.3.1

Atomic emission and absorption spectra result from the fact that electrons can move between energy levels when they have sufficient energy put in. They will then fall back, emitting a defined amount of energy as light. emission spectra result from electrons being excited by electricity or something, then emitting light as the electrons fall back. Absorption spectra result from electrons absorbing energy from electromagnetic radiation, and so effectively blocking it. In the emission case, there will be a series of thin bands representing the wavelengths of light produced, and for absorption, there will be a full spectrum with some lines cut out (the wavelengths that were absorbed).

13.3.2 : Bohrs postulates

Some stable orbits exists (assumed circular)
electrons absorbs/emits energy in changing orbits
Quantum condition (rules for changing orbit):

$mvr = n (\frac{h}{2\pi})$ , where n=1,2,3,...

This formula can be derived by equating Angular momentum with $n \overline{h}$ ( $\overline{h} = \frac{h}{2\pi}$ ). Historically it is based on expt discovery.

We'll call it +(1)+ later.

13.3.3

Describe the spectrum of atomic hydrogen and account for it, using Bohrs' model.

(This account should be slightly more detailed than that required by the IB.)

The charge in the nucleus = Ze

charge in electron = e

Assuming a circular path, centripetal force $= \frac{mv^2}{r}$ .

Since this force is supported by the electronic attraction, $\frac{mv^2}{r} = k \frac{(Ze)e}{r^2}$

Simplifying we obtain $r = k \frac{(Z)e^2}{mv^2}$ . Let's call it +(2)+

From above +(1)+, $mvr = n\overline{h}$ ,

thus rearranging we obtain $v = \frac{n\overline{h}}{mr}$ .

Plugging into +(2)+, simplify and you'll obtain a equation, which relates the orbital radius with (constant)n².

Energy of the electron $= KE + PE = \frac{1}{2}mv^2 + k \frac{Ze^2}{r}$ . Substitute by expression +(1)+, we'll arrive at E at nth shell $= \frac{(\mathrm{constant})}{r}$ at nth shell.

Since is proportional to n^2 , $E = \frac{\mathrm{constant}}{n^2}$ . (required by IB)

Knowing this and $E = \frac{hc}{\lambda}$ (Planck's equation), you'll also need to know the Rydberg eqn. This is shown in the syllabus, and can be easily derived from the above.

13.3.4 : Evaluate the success and limitations of Bohr's model

Success

Explains why atoms emit and successfully predict emission for Hydrogen.
Explains why atoms absorb
Ensures the stability of atoms
Predicts accurately the ionisation energy for Hydrogen.

Limitations

Not successful for multi-electron atoms.
Can't explain fine structures (emission lines existing as 2 or more close lines)
Can't explain bonding of atoms in molecules or solids & liquids
Can't explain different intensity of spectral lines.

13.4 Energy changes within nuclei

13.4.1 : Einstein's mass-energy equivalence

$\Delta E = \Delta m c^2$ .

is in joules, and in kilograms.

13.4.2 : term definitions

Unified mass unit - one twelfth the mass of a carbon 12 atom.

Mass defect - the amount of mass which is converted into energy in a nuclear reaction.

Binding energy - the energy equivalent of the difference in mass between the nucleus of an atom, and the masses of the individual protons and neutrons which make it up.

13.4.3

The binding energy can be calculated as described above, but finding the total mass defect between the individual nucleons, and the whole nucleus. The binding energy per nucleon is therefore, this divided by the number there are.

13.4.4

The graph of atomic number vs binding energy per nucleon runs from Z=2 increasing rapidly (with a 3 peaks I don't think we need to worry about) to about Z=20 where it runs relatively flat at around 8 Mev per nucleon then begins to drop off after Z = 60. The higher an element is (ie the more binding energy it has) the more stable it will be, and so the most stable elements are those around Z = 20.

13.4.5

Fission is the process by which an atom breaks up into smaller fragments. This is often caused by the addition of neutrons to the atoms, causing it to become unstable and eventually break up. This breaking up may, in some cases, produce more neutrons, and so these can then go on to produce more fission reactions, creating a chain reaction which perpetuates itself.

13.4.6

Fission is good because it provides a lot of energy form a source that is more viable long term than fossil fuels, and because it is relatively clean in terms of air pollution compared to fossil fuels. The down side is, it produces radioactive material which must be stored somewhere, and also, it can be dangerous if not controlled properly (meltdowns and stuff...)

13.4.7

Nuclear fusion occurs when two smaller nuclei fuse together to form one bigger, and more stable nucleus, and produce lots of energy in the process. Initiation of fusion requires a great deal of heat, because the nuclei must be given enough initial energy to overcome the coulomb repulsion between them as they approach. Energy calculations can be done using:

$\displaystyle E = mc^2$

when the masses of the different fragments are given.

13.5 Interaction of matter and energy

13.5.1

The explanation of the photoelectric effect is that the energy carried by light is broken into discrete units, the size of which depend on the frequency of the radiation. the energy carried in each 'photon' is defined as E = hf (plank's constant x frequency). The atoms require a certain amount of energy to release an electron, W_o = hf_o . where W_o is called the work function. if there is more energy than this, then that may be given to the electron in the form of kinetic energy, and so:

$\displaystyle E = hf = W_o + \frac{1}{2} m v_{max}^2$ .

13.5.2

The photoelectric effect can be measured by applying a stopping voltage in the opposite direction to the current induced by the photoelectron emission. As the frequency of the light is increased, more energy will be required to stop these electrons. If the frequency is decreased, however, there is eventually a point where no emissions occur, and so no voltage is required.

$\displaystyle hf = hf_o + eV_s$ , where V_s is the stopping voltage.

13.5.3

X-rays are produced by first placing an anode and cathode in a vacuum tube. Behind the anode is some type of photo-sensitive material, and between the two is a potential difference of about 150 000v. The cathode is heated to produce thermo-electrons. These electrons the accelerate towards the anode. When the electrons are deflected, by coming close to the nuclei, their kinetic energy changes. This change results in the production of an x-ray. Since the electrons can come as close or as far away from the nucleus, the x-ray spectrum is continuous, not discrete. There are, however, peaks caused by inner shell electrons being excited by small energy loss. These peaks occur on the left side of the curve, which is generally an inverted parabola type shape. There is a shortest wavelength possible for the x-rays due to the fact that when electrons lose all their kinetic energy, there is not way to make higher frequency waves.

13.5.4

When electrons lose some of their energy, x-rays are produced...the short wavelength limit is when they lose all of it so:

$\displaystyle eV = \frac{hc}{\lambda} = \frac{1}{2} mv^2$ and then $\lambda_{min} = \frac{hc}{eV}$ .

13.5.5

DeBroglie's equation is:

$\displaystyle \lambda = \frac{h}{p}$ .

Thus, we can see that all mass has a wave equivalent, and any wave has a mass equivalent.

13.5.6

The velocity of an electron can be found from:

$\displaystyle \frac{1}{2}mv^2 = eV$ ,

and this can then be used in the equation:

$\displaystyle mv=\frac{h}{\lambda}$ ,

to find the electron's wavelength. This can be seen/verified by the diffraction of electrons through thin crystals, showing that electrons have a wave nature.

13.6 Particle physics

13.6.1

Linear accelerators are designed based on a series of 'tubes' through which the particles are pulled, and then pushed by electric fields. The lengths of the tubes become longer and longer so the frequency with which the electric fields must oscillate are constant.

13.6.2

Circular particle accelerators work on the basis of magnetic fields making the particles rotate, and when they cross between the two Ds, they are accelerated between them by a electric field. The radius inside is defined by $r = \frac{mv}{Bq}$ , and so as the velocity increase, the magnetic field must be increased to keep the radius constant.

13.6.3

When a particle having been accelerated collides with a fixed target, both usually break up into smaller fragments. These can sometimes be identified in a cloud chamber.

13.6.4

Particle anti-particle pairs are only really produced from interactions involving great amounts of energy. Then two such particles collide, the completely annihilate, producing only energy.