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to Atomic and Nuclear Physics Welcome! All these slide presentations are at: http://www.hep.shef.ac.uk/Phil/PHY008.htm and also on FY website Phil Lightfoot, E47, (24533) p.k.lightfoot@shef.ac.uk

Most Important Thing !!!!!! I’m always available to help with any aspect of the course Stop me if you’re confused My contact details are on the top of your lecture notes

How to pass Physics !!!!!! Read 3 or 4 pages ahead in the notes so you are familiar with content of the next lecture and in the lecture ask about the bits you didn’t understand. In the lecture have a go at answering the questions – there’s no such thing as a stupid answer. Have a go at the questions from the problem classes. These are very similar to the exam questions and if you can do these then you’ll be fine. Just because your first homework is set for 27th April doesn’t mean you can’t get loads of feedback on how you’re doing. Ask questions, do problem class questions, have a go at questions in lectures. Come to see me whenever you don’t understand something. Don’t wait until just before the exam. Email me saying what you want to talk about and when you’re free and we’ll go somewhere until you’re happy.

Review of last lecture: The Photoelectric effect Scientist noted that when some metals were illuminated by a source of light, electrons were emitted from the metal with a certain kinetic energy. But the scientists were confused. The kinetic energy of the ejected photoelectrons was found to depend on the frequency of light and the type of metal used and not, as expected, on the brightness of the light source. Einstein offered an explanation in 1905 by proposing that light was not simply a wave but rather made up of tiny quanta or packets of energy, the energy of a single photon proportional to the frequency of light. They had a big problem thinking of it as a particle like a ball since then it would have to have weight and processes like interference and diffraction couldn’t be explained. What Einstein was saying was that we should actually think of light as a bunch of tiny packets of energy each with a small amount of energy. He even went further and said that each photon (packet of energy) would have an energy of : Where f is the frequency of the light and h is Plank’s constant (6.63 × 10-34 Js)

The Photoelectric effect Einstein proposed that there was a minimum energy E0 required to release a photoelectron from a metal. He called E0 the work function and suggested that this value was a constant for a particular metal, but was different for different metals. When a photon is absorbed within a metal, some of the photon’s energy will be used up in freeing a photoelectron from the metal, and if there is any energy remaining, then this will appear as kinetic energy of the ejected photoelectron. Where f is frequency of light and h is Plank’s constant (6.63 × 10-34 Js), E0 is the work function in joules.

The Compton effect In the last section we saw how Einstein imagined waves of light to be in fact tiny particles called photons in order to explain the photoelectric effect, and how he linked the energy of the photon to the frequency of the wave with the equation Where f is frequency of light and h is Plank’s constant (6.63 × 10-34 Js). But this isn’t Einstein’s most well known equation !!! What is ??? Where m is the mass of the photon and c is the speed of light In this equation he is defining a relationship between the energy of the photon of light and its ‘mass’ imagining it to a ball-like packet. So now we have two expressions for the energy of a photon. Let’s equate them to each other Momentum p is always equal to its mass multiplied by its velocity i.e. So and since therefore

The Compton effect Where p is the momentum in kgms-1, h is 6.63×10-34 Js, and λ is the wavelength in m. This is called the de Broglie equation after the person that did the maths based on Einstein’s equations. This is a very important equation because it links momentum of a photon to its wavelength. Many scientists questioned this interpretation. What was needed was an experiment to demonstrate the particle nature of photons of light. In 1923 Arthur Compton did this by setting up a collision between X-ray photons and electrons. relationship. http://faculty.gvsu.edu/majumdak/public_html/OnlineMaterials/ModPhys/QM/Compton/compton.html

The Compton effect The experiment showed that the X-ray photons and electrons behaved exactly like ball bearings colliding on a table top. Because the electron was scattered, the photon must have transferred both momentum and kinetic energy to it. This can only be explained by assuming that photons have momentum. But he observed something else. Before the collision the photon had one wavelength and after the collision its wavelength had increased. Remember that since if the photon loses energy then λ decreases Clearly the electron had been given energy, conservation of energy indicating that the scattered photon must therefore have lower energy than prior to the collision. The Compton effect is important because it demonstrates that light cannot be explained purely as a wave phenomenon, the classical theory of an electromagnetic wave scattered by charged particles unable to explain any shift in wavelength.

The Compton effect : an example Let’s imagine that we collide a photon of green light (λ = 530 nm) with a stationary electron. If after the collision the momentum is shared equally between the photon and the electron, what is the final wavelength of the photon and the velocity of the electron? What is initial momentum of a photon of green light? What is the mass of an electron? 9.1 × 10-31 kg What is the momentum of a stationary electron? zero If momentum is shared equally, how much does the photon end up with ? 0.625 × 10-27 kgms-1 What is the final wavelength of the photon ? What is the final velocity of the electron ?

de Broglie and matter waves So de Broglie had said that light, previously thought of as a wave could actually be thought of as a particle with momentum like a ball. The next obvious question of course was…. Could particles previously thought of as behaving like balls actually behave like waves in some situations??? This became known as wave-particle duality. To test this concept de Broglie needed to try to get matter to demonstrate wave-like properties such as diffraction or interference for example using a double slit apparatus. Interference Wave-like behaviour

de Broglie and matter waves What would you expect if you fired two Uzis through two holes in a wall ? Would the bullets act like particles or waves? Let’s repeat the experiment but instead of using bullets lets use electrons As expected the bullets from different guns don’t interfere with each other and you just get bullet holes in two specific places. Big heavy bullets are acting like particles If they had instead acted like waves what would the bullet hole distribution on the wall have looked like ??

de Broglie and matter waves When we repeat this experiment using electrons we find amazingly that we get interference patterns on the far wall even though we think of electrons as little balls!! The only way that this can be explained is if the electrons are acting like waves (called ‘matter waves’) and constructively and destructively interfering. http://chaos.nus.edu.sg/simulations/Modern%20Physics/Interference/interference.html

de Broglie and matter waves Let’s try to work out why electrons can act like waves but bullets do not …. Earlier we derived the de Broglie equation where p is the momentum in kgms-1, h is 6.63×10-34 Js, and λ is the wavelength in m. Momentum of a particle is defined as its velocity multiplied by its mass where m is the mass and u is the velocity. The kinetic energy of a particle is given as E so: So by rearranging we find And therefore . Finally we have the expression :

de Broglie and matter waves p is the momentum in kgms-1, m is in kg, E is in joules, h is 6.63×10-34 Js, and λ is the wavelength in m. From double slit experiments using light we know that we only see clear evidence of interference (i.e. those pretty interference fringes) if the spacing of the slits is about the same as the wavelength  of the light. It is therefore sensible to select matter waves of similar wavelength to the slit spacing. This caused de Broglie a great deal of difficultly. Let’s see why using an example. What is the momentum of an electron of mass 9.110-31 kg moving with velocity u = 4.68107 m s-1 ?? What is its wavelength according to the de Broglie equation ?? . This is really really small. He couldn’t find or make a slit that narrow so he couldn’t demonstrate interference.

de Broglie and matter waves The best he could do was select a very thin crystal of carbon atoms close to this spacing. He then shone an electron beam at it, the regular spacing of atoms forming a kind of diffraction grating producing rings on a distant screen. This diffraction pattern is of very similar appearance to that seen from a light source of similar wavelength, as shown in the figure to the right. . This was proof that electrons were acting like waves.

de Broglie and matter waves p is the momentum in kgms-1, m is in kg, E is in joules, h is 6.63×10-34 Js, and λ is the wavelength in m. So why don’t we see diffraction effects when a person collides with a metal grating?! People are heavier than electrons and therefore they have a higher momentum. Looking at the equation above a big momentum p means that λ will be tiny. Remember that in order to see diffraction effects the wavelength should be approximately the same as the slit separation. But the wavelength λ of the matter wave is incredibly small and certainly far smaller than the smallest slit separations. Question : How slow does a 2000 kg elephant have to move in order to have the same momentum as an electron travelling at 2×107 ms-1 ?? . This also explains why we see fringes in the double slit experiment using electrons but not bullets !!!

Rutherford: The gold foil scattering experiment In 1904, Thomson considered atomic structure to be represented by a ‘plum pudding’ in so far as the atom was made up of negatively charged electrons surrounded by a soup of positive charge to keep the overall charge neutral. To find out for sure, in 1909 Rutherford fired positively charged helium ions (sometimes called alpha particles) with high energy at gold atoms in a thin gold film. Since it was believed that positive and negative charges were spread evenly within the atom and that therefore only weak electric forces would be exerted on the helium ions passing through the thin foil, he expected to find that most ions travelled straight through the foil with no deviation. .

Rutherford: The gold foil scattering experiment What he found, to great surprise, was that whilst most passed straight through the foil, a small percentage (about 1 in 10000) were deflected at very large angles and some even bounced back toward the ion source. What did this say about the atomic structure of gold ?? Because helium ions are about 8000 times the mass of an electron and impacted the foil at very high velocities, it was clear that very strong forces were necessary to deflect these particles. (Imagine firing bullets at soup. Even if just one ricocheted back it would be surprising!!) http://serc.carleton.edu/sp/compadre/interactive/examples/19267.html . http://webphysics.davidson.edu/Applets/pqp_preview/contents/pqp_errata/cd_errata_fixes/section4_7.html http://galileoandeinstein.physics.virginia.edu/more_stuff/Applets/rutherford/rutherford.html http://galileoandeinstein.physics.virginia.edu/more_stuff/Applets/rutherford/rutherford2.html

Rutherford: The gold foil scattering experiment The positive helium ions had clearly been repelled by an incredibly large positive charge within the atom, this charge concentrated in a dense region also containing most of the mass. This work led in 1913 to Rutherford declaring the atom to contain a very small nucleus of high positive charge (equal to the number of electrons in order to maintain neutrality) and to be similar to the ‘solar-system-like’ model, in which a positively charged nucleus is surrounded by an equal number of electrons in orbital shells. We now know that the diameter of the nucleus ranges from 2 to 9 × 10-15 m from hydrogen to uranium, and the diameter of the nucleus to be almost 10-5 the diameter of the atom. .

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