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30, 31, 32 Tips
Physics 30 & 31 Tips
| Question | Answer |
|---|---|
| Continuous Spectrum | all objects emit radiation. The continuous spectrum is due to the radiation that is emitted by the entire collection of atoms that are bound together to form the solid. Visible part of the spectrum includes the entire range of visible wavelengths. |
| Line Spectra | Energized gas atoms emit light at specific wavelengths as electrons fall between discrete energy levels, creating a line spectrum. A diffraction grating separates the wavelengths by bending each color at a different angle. |
| Balmer | Balmer developed an equation that predicts the wavelengths of four visible lines, Emission equation where Z is the atomic number and R is a constant. The electron falls from ni to nf, releasing a photon. |
| Lyman Series | (nf=1) Ultraviolet (very short wavelengths) Example: transitions like 2 → 1, 3 → 1 |
| Balmer Series | (nf=2) Visible light (this is what Balmer studied) Example: 3 → 2, 4 → 2 These are the colored lines you can actually see |
| Paschen Series | (nf=3) Infrared (longer wavelengths, not visible) |
| Bohr’s Model of the Hydrogen Atom (Part One) | The electron travels in circular orbits about the positively charged nucleus. However, only certain orbits are allowed. The electron does not radiate energy when it is in one of these orbits. Net force equals mass * centripetal acceleration. |
| Can Bohr's Model be Applied to Lighter Atoms? | Yes but only after you ionize the atom and leave a single electron orbiting the nucleus. One electron needs to be removed from the helium atom to turn it into helium plus one and two electrons need to be removed from lithium to become lithium plus two. |
| Bohr’s Model of the Hydrogen Atom (Part Two) | If an electron falls from one orbit, also known as an energy level, to another, it loses energy in the form of a photon of light. The energy of the photon equals the difference between the energy of the orbits. This is called the emission process. |
| Bohr’s Model of the Hydrogen Atom (Part Three) | The orbits have angular momentum (L) given by a specific equation where rn would equal the radius of the orbit. Angular momentum can only take specific (quantized) values, not anything continuous. |
| Bohr’s Model of the Hydrogen Atom (Part Four) | A hydrogen atom can absorb only those photons of light which will cause the electron to jump from a lower level to a higher level. Thus, the energy of the photon must equal the difference in the energy between the two levels. |
| Requirements for Absorption | If an electron is in the ground state (n=1) and a photon has exactly the energy needed to jump to n = 2, the atom absorbs it. If the photon has slightly more or less energy than required, it won’t be absorbed at all. |
| More Requirements for Absorption | A photon cannot be absorbed by a given atom unless the atom has TWO energy levels where the difference between their energies is exactly equal to the energy of the photon. |
| Example Image | In the picture, the dark band or missing color is due to the fact that photons have been absorbed by a gas with two energy levels where their difference is exactly equal to the energy of the missing photon. |
| Atomic number | the number of protons found in the nucleus of an atom of a chemical element |
| Determine the energy required to ionize an electron located in the ground state of the lithium 2+ ion? | First you have to take two e- away to treat it with Bohr model. Ground state means n = 1, you want to move e- from 1 to the highest level, infinity. Use the En energy equation with Z = 3. Do Einfiinity - E1, infinity cancels out, solve. |
| X-Ray Purpose | In an X-ray tube, a low-voltage source heats a filament to release electrons, while a high-voltage source accelerates them toward a metal target. When the fast-moving electrons hit the target, X-rays are produced. |
| Breaking Radiation (Bremsstrahlung) | When fast-moving electrons hit the target metal, they suddenly slow down and release photons with many different energies. This produces Bremsstrahlung (“breaking radiation”), which creates the continuous background spectrum of X-rays. |
| Threshold wavelength (λ0) | The strongest photon is the one that corresponds to the smallest wavelength, known as the threshold wavelength (λ0). That photon is released when the bombarding electron loses all of its energy at once upon contact with the target metal. |
| Threshold wavelength (λ0) Formula - NOT GIVEN | λ0 = hc/qV |
| Why is Q delta V positive? | The qdeltaV quantity is pos, because the charge of the e- is neg and with the neg sign in front of it, it becomes pos. The delta V itself is pos because the final potential near the target metal is higher than the initial potential near the filament. |
| Characteristic X-Rays (Step Two of X-Ray Production) | Characteristic X-rays are produced when bombarding electrons knock out inner-shell electrons from the target metal atoms. Electrons from higher energy levels then fall into the vacancy, releasing X-rays with specific energies characteristic of the metal. |
| K-alpha (Characteristic X-Rays): | electron falls from level 2 → 1 (more common → bigger peak) |
| K-beta (Characteristic X-Rays): | electron falls from level 3 → 1 (less common → smaller peak) |
| The spectrum has: | A smooth background (from braking radiation) Sharp spikes (from characteristic X-rays) |
| Increasing voltage: | Makes electrons more energetic Shifts the threshold wavelength to shorter values (left side of the graph) |
| Determine the minimum accelerating voltage required to give an electron enough energy to produce the strongest X-ray? | Strongest X-ray is the one with the threshold wavelength, produced when the accelerated electron loses all of its energy upon collision with the target metal. Do qV = hc/λ, solve for V. |
| How to determine how many possible different photons are emitted by an atom? | Draw out all of the orbits, draw the arrows for each possible transition down, remove duplicated photons and keep one only, the leftover is your answer. |
| How to find the longest wavelength that can be emitted? | Use Eph = hc/lambda, do Eni - Enf first, then find the smallest difference between n levels and insert that into the equation. |
| As the values of n increases... | The difference in energy between the two adjacent levels decreases as a result, so the photon that emits the longest wavelength is the one that corresponds to a transition made between the largest energy levels. |
| Important Equation to know for shortest wavelength | Eph = 13.6 Z^2 (1/nf^2 - 1/ni^2) |
| Common Atomic Numbers | Hydrogen = 1, Helium = 2, Lithium = 3, Beryllium = 4 |
| How to solve for the shortest wavelength of an unknown series? | ni would be equal to infinity, set up hc/lambda = E infinity - En for lithium, then beryllium (unknown). Set up a ratio between the equations, solve for lambda 2 |
| How to solve for the wavelength of an electron that loses 60% of its energy to form a X-ray photon? | Use the qdeltaV = KEf, Kei would equal 0. Then multiply the Kef amount by 0.60. Input that value into Eph = hc/lamdba, then solve. |
| Structure of the Nucleus | the structure of the nucleus contains protons and neutrons. These particles are called nucleons. Electrons are NOT considered nucleons. |
| Neutron vs. proton vs. electrons | The neutron is electrically neutral, while the proton carries a single electric charge. The electron, though not associated with nucleons, it carries a negative charge and has a charge magnitude that is the same as a proton’s. |
| The nucleus of a chemical element is designated by... | A Z (X) X is the chemical symbol of the element, Z is the atomic number, and A is the atomic mass number, which is neutrons and protons combined. |
| Radius Equation for a Nucleus | r = (1.2*10^15 m)(A ^1/3) Radius increases with mass number (A) |
| Isotopes | the nuclei of a given element all have the same number of protons but a different number of neutrons → these are known as isotopes. The number of neutrons is equal to the difference between the atomic mass number A and the atomic number Z |
| The Strong Nuclear Force and the Stability of the Nucleus | the nucleus is held together by an attractive force which acts between all nucleons, protons, and neutrons alike (can exist between n-n, a n-p, and p-p). Prevents the nucleus from breaking to electrostatic force which exists between the protons. |
| How is the nuclear force short-range? | Strong nuclear force is a short range. This force is essentially 0 if the separation distance between nucleons is greater than 10^-15 m. IN ORDER FOR A NUCLEUS TO BECOME STABLE, THE STRONG NUCLEAR FORCE MUST BE STRONGER THAN THE ELECTROSTATIC FORCES. |
| Long Range Repulsion Vs. Short Range Nuclear Force | As nuclei gets heavier, extra neutrons are needed to balance the increasing repulsion between protons through the strong nuclear force. After a certain size, long-range repulsion is too great for the short-range strong force to hold the nucleus together. |
| radioactivity | When the nucleus becomes unstable and spontaneously breaks apart, this process is known as radioactivity. |
| The Mass Defect of the Nucleus and Nuclear Binding Energy | the total mass of the nucleus is always less than the sum of the masses of the protons and neutrons of which it is composed. The energy equivalent of the mass difference (delta m) is the binding energy of the nucleus. |
| Binding Energy (Simple) | the binding energy is the energy used by the nucleus to keep the nucleons together as one unit. AND it is the energy that must be added to the nucleus to break it apart into separate neutrons and protons. |
| Total Binding Energy | Total binding energy is the energy produced from the mass defect — the difference between the mass of the separate nucleons and the actual mass of the nucleus — and it holds the nucleus together. |
| Average Binding Energy | obtained by dividing the total binding energy by the total number of nucleons. If you see the average binding energy slowly decreasing, that is because electrostatic forces are breaking it apart. |
| Alternate Equation for Mass Defect | delta m = m nucleons - m nucleus |
| What does mass of the nucleus equal? | M nucleus = atomic mass - (electron mass * # of electrons) |
| What does the mass of the nucleons equal? | (# of protons * proton mass) + (# of neutrons * neutron mass) |
| If the question is asking for the average binding energy per nucleon... | All you need to do is divide the total binding energy by the total number of nucleons. |
| Radioactive Decay | Henry Becquerel was the first to make the discovery of radioactivity. The source of radioactivity is the nucleus. Certain isotopes are not stable under the action of the nuclear force and they decay with the emission of some form of radiation. |
| What conservation laws do we focus on for radioactive decay? | The mass-energy conservation law, the linear-momentum conversation law, the angular-momentum conservation law, and the electric-charge conservation law. |
| What is alpha (α) radiation? | Alpha radiation consists of 2 protons and 2 neutrons (a helium nucleus). It has a +2 charge, low penetration, and is stopped by paper or skin. |
| Beta Decay | Beta decay can occur by emitting either an electron (β⁻) or a positron (β⁺), along with a neutrino or antineutrino. The emitted electron or positron is created inside the nucleus and is not an orbital electron. |
| Beta-minus (β⁻) decay | This happens when a nucleus has too many neutrons. A neutron converts into a proton: n → p + e^- + ve What’s emitted: an electron (e^-) → the beta particle an antineutrino (ve) |
| Beta-plus (β⁺) decay (positron emission) | This happens when a nucleus has too many protons. A proton converts into a neutron: p → n + e^+ + ve What’s emitted: a positron (e^+) a neutrino (ve) |
| Gamma Decay | occurs when a nucleus in an excited state drops to a lower energy state. A photon, gamma ray (y), is emitted, analogous to emission. The possible energy levels of a nucleus are so far apart and that’s the reason why high-energy photons are released. |
| Why does gamma decay have no change in nucleon number or atomic number? | Because it's a photon that carries no electric charge or rest mass. |
| Radioactive Decay and Activity | Radioactive nuclei decay randomly over time rather than all at once. The number of decays in a given time interval is proportional to both the length of the interval and the number of undecayed nuclei remaining. |
| Radioactive Decay and Activity (Simple) | Radioactive decay happens randomly, but the more nuclei you have and the longer the time interval, the more decays occur. So, the number of decays is proportional to both the number of remaining nuclei and the time elapsed. |
| Lambda in Radioactive Decay and Activity | Lambda is a constant of proportionality, which differs from one isotope to another. Also, a negative sign is used because the number of radioactive isotopes is constantly decreasing. |
| N tip! | N0 = number of nuclei present in the sample at t = 0 N = number of remaining radioactive nuclei |
| Activity (A) | Number of decays per unit time. A = the activity at a given time (the number of decays per second at that moment) A0 = the initial activity (the activity at time t=0) |
| SI unit for Activity | SI unit is Bq. The Curie unit, which is not an SI unit, can also be used, and it’s named after Marie Curie and her husband, Pierre, who isolated two previously unknown radioactive elements. |
| Radioactive Half Life | The half-life (T ½) of an isotope is the time required for half the radioactive nuclei present in the sample to decay. |
| How to determine the age of the artifact for a dead tree? | You would use the activity equation, A would be your decays per minute, convert that to decays per second, solve for Ao next, by multiplying your g of pure carbon by (6.023*10^23/12g)(1/10^12). Get No, put into Ao equation. Put into main equation, solve |
| Seconds to years conversion | Seconds / (60*60*24*365) |
| If they give you the activity of one gram of fresh carbon, but the question asks you to find the age of a sample of 10 g of pure carbon, then | Ao = 0.23 Bq Ao/10 g = 0.23 Bq * 10.0 g -> 2.3 Bq |
| If they don't give you the lambda constant, then... | You can find it using the half-life equation. |
| Grams Tip | If you're trying to do calculations with No, always multiply by Avogadro's number and divide by molar mass |
| Nuclear Reactions and Transmutation of Elements: | a nuclear reaction occurs when two nuclei collide and two (or more nuclei) are produced. When a nucleus undergoes a nuclear reaction, the daughter nucleus or nuclei formed are completely different from the parent nucleus. |
| General Reaction of Nuclear Reaction | x+X→y+Y x = incoming (bombarding) particle X = target nucleus (the one being hit) y = outgoing small particle (like a neutron, proton, alpha, or photon) Y = final (recoil) nucleus after the reaction |
| Short Form of Nuclear Reaction | X(x,y)Y This reads as: "Target nucleus X is hit by particle x, emits particle y, and becomes Y." |
| 1 1 H + 23 11 Na → 20 10 Ne + 4 2 He Explain all parts of this | 1 1 H + 23 11 Na → 20 10 Ne + 4 2 He 1 1 H = proton → this is x 23 11 Na = sodium nucleus → this is X 4 2 He = alpha particle → this is y 20 10 Ne = neon nucleus → this is Y |
| Short Form of Example Reaction | 23 11 Na(p,α)20 10 Ne Sodium-23 is bombarded by a proton, emits an alpha particle, and turns into Neon-20. |
| Tip about Nuclear Reaction Conservation | in nuclear reactions, the mass energy is conserved, the linear momentum is conserved, and the angular momentum is conserved. In addition, both electric charge and nucleon number are conserved. |
| There are two major nuclear reactions → | nuclear fission and nuclear fusion. |
| Nuclear Fission | happens spontaneously for heavy, unstable nuclei. When fission occurs, a massive nucleus divides into two fragments. |
| Nuclear Fusion | two low-mass nuclei combine to form a more massive nucleus. This process is not spontaneous. In both cases, the binding energy per nucleon is greater for the products than the reactants, meaning there is energy released in both processes. |
| Nuclear Fission Example | When a uranium-235 nucleus absorbs a slow-moving neutron, it becomes unstable uranium-236, releasing energy and additional neutrons. These emitted neutrons can then trigger further fission reactions, creating a self-sustaining chain reaction. |
| What happens to the missing mass in nuclear fission? | It can be seen that the sum of rest masses of reactants is greater than the sum of the rest masses of the products. The missing mass is converted to energy -> E = mc^2 |
| How to determine the energy of a fission reaction? | Remember that the # of orbital e- on the reactant side will always be different from the product side. The lone e just chilling are from the nucleus, WILL NOT BE COUNTED. First find m nucleus for reactants, then mreactants, then do the same for products. |
| How to determine the energy of a fission reaction? (Cont'd) | Continue finding the nucleus for all atoms until you have a definite value for both m reactants and m products. For products, you're adding both neutrons and electrons to it. Finally, do delta m = m reactants - m products, solve, multiply by 931.5 for E. |
| Fission ONLY occurs in certain elements… | fission only occurs in certain elements that have a large nucleon number, for example, 235-U and 239-Pu. Both of these nuclei may fission if struck by a slow moving neutron. |
| In order for fusion to happen... | it is necessary for the nuclei to collide while travelling at very high speeds. Such velocities can be achieved if the nuclei is part of a hot gas in which the temperature approximates 100 million degrees Kelvin. |
| Can Nuclear Fusion happen on Earth? | No, because in order for two positively charged nuclei to fuse, they need to overcome the strong barrier caused by repulsive electrostatic forces. Reactions like this commonly happen in stars like the sun. |
| When the reaction doesn't emit electrons or positrons, you can... | Just use the atomic masses to find m reactants and m products. Remember to add any neutrons you have though. |
| Determine the amount of an atom that would be consumed everyday of the output from the plant is to be 100,000 Watts and the plant is 33.3%? | Poutput/Pinput = 0.333, so 100,000 * 3 = 300,000 W. To find energy, you can do P * t, convert to MeV to J, then multiply mass of atom over energy produced, turn to Kg. |
| How to find the amount of energy released over a half-life? | Find Etotal and multiply by No/2 |
| How to find the speed of an alpha particle? | Find total energy, convert to J, set that energy equal to 1/2mv^2. Convert the given mass value to kg, solve for v |
Created by:
smurtab