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All Physics 29 Tips
| Question | Answer |
|---|---|
| Thermal Radiation | at any temperature, an object emits electromagnetic radiation called thermal radiation. Thermal radiation consists of a continuous spectrum, encompassing wavelengths from the visible, ultraviolet, and infrared portions of the spectrum. |
| The emission of radiation can be seen by the fact that objects glow when we increase their temperature, such as a red hot stove. | Temperature around 1000K → red light Temperatures above 2000K → yellow or white-like colors, like the filament of a light bulb |
| The fact that color changes with increasing temperature means... | the frequency of the radiation increases with increasing temperature. |
| Electromagnetic Radiation | from the classical point of view, electromagnetic radiation is generated by accelerating particles. The distribution of accelerations produces a continuous spectrum. |
| Blackbody | an ideal, hot, dense, object. A blackbody is a body that would absorb all radiation falling on it. |
| quantized | means that something can only take on certain discrete values rather than any value continuously. Max Planck said molecules in a heated object can vibrate only with discrete amounts of energy -> E = nhf |
| Tip about n (the photon): | a photon is not a particle; instead it is a quantity of energy that is equal to hf. |
| Photoelectric effect | light of a high enough f shines on a metal surface and ejects e-, producing a photocurrent. Shows that electromagnetic energy is quantized into photons, each with E=hf, where increasing light intensity increases the # of photons but not the energy. |
| Photoelectric effect EXPLAINED | when the intensity of light is increased, the number of photons in the beam increases, but the energy of each photon remains the SAME because the energy depends only upon the frequency of the wave. |
| Work Function | minimum amount of energy required to move an electron from the surface of metal. The energy of each photon is going to split between the work function and the kinetic energy carried by the photoelectrons. |
| When the maximum kinetic energy is equal to zero... | The work function equals hf0, where f0 is the threshold frequency and lamba0 equals the threshold wavelength. |
| Verified Predictions for the Photoelectric Effect (1) | The number of electrons emitted per second (photocurrent) increases as the intensity increases. The maximum kinetic energy of the electrons is not affected by the intensity of the light. |
| Verified Predictions for the Photoelectric Effect (2) | The maximum kinetic energy of the photoelectrons is increased when the energy of the photons is increased. |
| Verified Predictions for the Photoelectric Effect (3) | Below a certain frequency, called threshold frequency, no electrons are ejected. |
| How to find the work function (Wo) for a surface? | If you're given velocity max, you can input that into 1/2mv^2 for KEmax. Convert into eV if needed. Then if you have a wavelength, you can do hc/wavelength and solve for the rest of the equation. Convert this into eV too. |
| eV to J | Multiply by 1.6 * 10^-19 |
| J to eV | Divide by 1.6 * 10^-19 |
| How to find the cutoff frequency (fo)? | You can use the Wo = hfo equation (NOT GIVEN) |
| Tip about cutoff frequency (fo) | When you evaluate the frequency, you would use the work function in joules |
| Tip about hf | hc/lambda can always be swapped for hf |
| eVo Tip | can be swapped for qVo |
| Stopping Potential | minimum negative potential required to stop the most energetic (KEmax) photoelectrons. AKA the smallest potential difference between the plates that brings the photoelectrons to a stop just before they reach the negatively charged plate. |
| Stopping Potential Experiment | battery is reversed so the emitted photoe- are opposed by an increasing electric potential. Voltage is increased until the photocurrent becomes 0, and this stopping potential is used to determine the max KE of the e- and the work function of the material. |
| Pair Production | Pair production occurs when a high-energy gamma-ray photon passes near an atomic nucleus and transforms into an electron and a positron. The electron and positron have equal mass and equal but opposite electric charges. |
| Why Do Electrons and Positrons have opposite charges? | The electric charge must be conserved, which is why the produced pair has opposite electric charges → so the charge of the products is as neutral as the photon. |
| Conversation During Pair Production | charge, momentum, and energy must all be conserved. The photon must have at least enough energy to equal the rest mass of the e- and positron, while any extra energy becomes kinetic energy, and a nearby nucleus helps conserve momentum during the process. |
| Antiparticle | In pair production, the antiparticle is the positron, which is the antiparticle of the electron. |
| The products MUST have a momentum equal to | the momentum of the photon for the pair to be produced. |
| Pair Production (Simple) | Photon + near nucleus -> electron & positron |
| Pair annihilation | a particle and its antiparticle, like a e- and a positron, collide and destroy each other, converting their mass into energy in the form of 2 photons. 2 photons are produced so that momentum is conserved, with the photons moving in opposite directions. |
| Pair Production Vs. Pair Annihilation | Pair production is when a high-energy photon creates a particle–antiparticle pair, turning energy to matter. Pair annihilation is the reverse process where they destroy each other and produce photons, turning matter to energy |
| Pair Annihilation Conservation | In pair annihilation, a particle and its antiparticle (with equal and opposite charge) cancel each other, producing neutral photons so charge is conserved. At least two photons are created moving in opposite directions to conserve momentum. |
| What is the minimum energy of a photon required to produce a proton-antiproton pair? | Since the energy in minimum, the pair must be created at rest with no extra energy. Ephoton = moc^2 (p+) + moc^2 (p-). So Ephoton - 2moc^2, solve and convert to eV. |
| Important Equation That's Not Included | Ephoton = moc^2 (for the proton) + moc^2 (for the anti-proton) |
| Tip about minimum energy | Means the pair must be created at REST, you're using rest mass energy |
| Wave Particle Duality | Light (an electromagnetic wave) sometimes behaves like it’s made of particles called photons, each carrying energy hf. Based on that idea, de Broglie suggested that matter particles (like electrons) can also behave like waves, not just particles. |
| Wave Particle Duality (Simplified) | light can act like particles, and particles can act like waves. |
| Reminder about photons | photons are a quantity of energy that equals hf, is not a real particle, and it does not have a mass. |
| Tip about momentum | The momentum (p) for particles and photons will be different, because photons do not have mass. |
| Uncertainty Principle | Heisenberg was responsible for a concept called the uncertainty principle. States that it is fundamentally impossible to make simultaneous measurements of a particle’s position and velocity with arbitrarily large accuracy (100% accuracy). |
| Can uncertainty principle be fixed? | Sometimes, these uncertainties can be reduced by more precise instruments, but usually not. |
| delta py and delta y in uncertainty | uncertainty in momentum along the y direction, uncertainty in position along the y direction |
| How to compute minimum uncertainty? | Solve for uncertainty in measurement of speed first by doing delta V / V -> delta V = 0.05V, then use the PV > h/4pi equation. Remember delta P can equal mdeltaV. Mass must be converted to kg, put that into equation and solve for JUST delta y. Use = |
| Mass of diatomic molecules | Would be 2 times regular mass, for example 2 * 14 g/mol. You can convert that to kg by dividing by avogadro's number then dividing by 1,000. |
| Compton Scattering | Arthur Compton shot a beam of X-rays at graphite. Found that the wavelength of the X-rays was longer, and therefore, the energy was smaller. Also found that the amount by which the energy was reduced was dependent on the angle of ray scattering. |
| Compton Scattering Additional | The incident photon collides with the stationary electron, giving up some of its energy to the electron, which recoils in one direction while the photon scatters in another, keeping the rest of the energy. |
| Compton Wavelength | Part of the equation that's h/moc, equal to 0.00243 nm. The equation as a whole is meant to help you predict the wavelength change after collision. |
| How to find the wavelength of a scattered photon? | Use the Compton scattering equation and find lambda prime. When a photon is reflected directly back from another, the angle would be 180. |
| Wavelength prime and Wavelength | wavelength of the scattered photon is lambda prime, regular lambda is wavelength of incident photon. Theta is the angle made between the scattered photon and the incident photon |
| How to solve for energy of recoil photon? | You can do Ke = Eo - E', using hc/lamdba equation for both with their respective wavelengths. |
| How to solve for the maximum wavelength of light incident on platinum that releases photoelectrons? | If you already have fo, you can just do fo = c/lambda |
| Given energy, how can you solve for stopping potential that can arrest photoelectric current? | Remember KEmax = qVo. Just divide KEmax/q, substitute the KEmax to eV but the q must just be one electron. |
| If you start with a proton-anti proton pair... | It's likely to be proton annihilation, cause that pair is going to form photons later. |
| If a proton-antiproton pair starts at rest, what are the energy and frequency of each of the emitted photons? | Since both particles are at rest: Total momentum = 0 → the two photons must travel in opposite directions with equal energy Total energy = rest mass energy of both particles, each photon gets half of 2moc^2, so just moc^2. Convert to J for E = hf |
| How to find the frequency of a photon that has the same momentum as a neutron? | Neutrons have p = mv and photons have p = hf/c, so set the two equation to each other. |
Created by:
smurtab