Loading studyCave
Preparing your tutoring experience...
Preparing your tutoring experience...
Master the evolution of the universe, atomic structure from Thomson to Schrödinger, and nuclear physics including the Standard Model with clear theory and exam-ready insights.
The Big Bang theory proposes that the universe began approximately 13.8 billion years ago from an extremely hot, dense state and has been expanding ever since. Three key pieces of observational evidence support this theory.
A galaxy is observed to be receding at 5250 km/s. Using Hubble's Law with , calculate: (a) the distance to the galaxy in Mpc, (b) estimate the age of the universe assuming constant expansion rate.
Given:
(a) Distance to galaxy:
Using Hubble's Law:
(b) Age of universe estimate:
The Hubble time gives approximate age:
Convert units:
Converting to years:
Close to the accepted age of 13.8 billion years!
Stellar spectra are produced when starlight is dispersed by a prism or diffraction grating. Analyzing these spectra reveals three crucial properties of stars: surface temperature, chemical composition, and radial velocity.
A stellar spectrum appears as a continuous rainbow of colors (continuous spectrum from the hot photosphere) crossed by dark absorption lines (Fraunhofer lines). These absorption lines form when light from the hot interior passes through the cooler outer atmosphere, where atoms absorb specific wavelengths corresponding to their electron transitions.
A star's spectrum shows peak intensity at 580 nm. A hydrogen absorption line normally at 656 nm appears at 658 nm. Calculate: (a) the star's surface temperature, (b) the star's radial velocity.
Given:
(a) Surface temperature:
Using Wien's Law:
Slightly cooler than the Sun (5800 K), appearing yellow-orange
(b) Radial velocity:
Wavelength shift:
Since shift is toward longer wavelength (red), star is receding.
Star is moving away from Earth at 915 km/s
The Hertzsprung-Russell diagram is one of astronomy's most important tools, plotting stellar luminosity against surface temperature (or spectral class). This reveals distinct groupings of stars and allows astronomers to infer stellar properties, ages, and evolutionary stages.
The H-R diagram reveals that stellar properties are not randomly distributed. The distinct groupings correspond to different stages of stellar evolution. A star's position changes as it ages: main sequence stars eventually become red giants or supergiants, and low-mass stars end as white dwarfs. The diagram is essentially a stellar evolutionary roadmap.
Star A has surface temperature 6000 K and luminosity 1 L☉. Star B has surface temperature 3000 K and luminosity 100 L☉. Identify which region of the H-R diagram each star belongs to and explain the physical reason for Star B's high luminosity despite low temperature.
Star A (T = 6000 K, L = 1 L☉):
Surface temperature similar to the Sun (5800 K) and luminosity equal to the Sun.
Classification: Main Sequence star
Star A is a solar-type star actively fusing hydrogen in its core, positioned in the middle of the main sequence.
Star B (T = 3000 K, L = 100 L☉):
Low surface temperature (cool, red) but very high luminosity.
Classification: Red Giant
Physical explanation using Stefan-Boltzmann Law:
For Star B to be 100 times more luminous than Star A while being cooler, it must have a much larger radius:
Star B has about 40 times the radius of Star A. Its enormous surface area more than compensates for its lower surface temperature, making it highly luminous.
Stellar nucleosynthesis is the process by which stars create heavier elements from lighter ones through nuclear fusion. Different fusion processes occur at different stages of stellar evolution, depending on the star's mass and core temperature.
In the proton-proton chain, four hydrogen nuclei (total mass 6.693 × 10⁻²⁷ kg) fuse to form one helium nucleus (mass 6.645 × 10⁻²⁷ kg). Calculate: (a) the mass defect, (b) the energy released per fusion reaction, (c) the energy released if 1 kg of hydrogen is completely converted.
Given:
(a) Mass defect:
About 0.7% of the original mass is converted to energy
(b) Energy per reaction:
This seems tiny, but remember this is per fusion event!
(c) Energy from 1 kg hydrogen:
Number of reactions:
Total energy:
Equivalent to about 150,000 tons of TNT! The Sun converts 600 million tons of hydrogen every second.
In 1897, J.J. Thomson conducted experiments with cathode rays to determine their nature. His work proved that cathode rays are streams of negatively charged particles (electrons) and allowed him to measure their charge-to-mass ratio, fundamentally changing our understanding of matter.
Thomson concluded that cathode rays consisted of negatively charged particles with very small mass - much smaller than any atom. He called these particles "corpuscles," but they became known as electrons. This discovery proved that atoms were not indivisible fundamental particles, but contained smaller components. Thomson proposed the "plum pudding model" where electrons were embedded in a diffuse positive charge, like raisins in pudding.
In Thomson's experiment, perpendicular electric and magnetic fields are applied to a cathode ray beam. The electric field strength is 2.0 × 10⁴ N/C and the magnetic field is 5.0 × 10⁻³ T. Calculate: (a) the velocity of the electrons when the beam is undeflected, (b) explain why this balance method allows velocity determination.
Given:
(a) Electron velocity:
When undeflected, electric force equals magnetic force:
The charge q cancels:
About 1.3% the speed of light
(b) Why this method works:
This balance method is powerful because the velocity can be determined without knowing the charge or mass of the particles. When forces balance, the equation v = E/B depends only on the known field strengths, not on particle properties.
Once velocity is known, Thomson could measure the deflection with only the electric field active. The deflection depends on q/m, allowing him to calculate the charge-to-mass ratio. This two-step process (balance to find v, then deflection to find q/m) was crucial to Thomson's discovery.
In 1909, Robert Millikan performed his famous oil drop experiment to measure the elementary charge. By carefully balancing gravitational and electric forces on charged oil droplets, he proved that electric charge is quantized and determined the fundamental unit of charge.
An oil droplet of mass 3.2 × 10⁻¹⁵ kg is suspended between parallel plates separated by 5.0 mm with a potential difference of 400 V. Calculate: (a) the electric field strength, (b) the charge on the droplet, (c) how many excess electrons the droplet carries.
Given:
(a) Electric field strength:
For uniform field between parallel plates:
(b) Charge on droplet:
At equilibrium:
(c) Number of excess electrons:
The droplet carries 2 excess electrons (small deviation due to rounding in measurements)
In 1909, Ernest Rutherford, along with Hans Geiger and Ernest Marsden, performed the gold foil experiment that revolutionized atomic theory. By firing alpha particles at thin gold foil, they discovered that atoms have a tiny, dense, positively charged nucleus.
Estimate the distance of closest approach when a 5.0 MeV alpha particle (charge +2e) approaches a gold nucleus (charge +79e) head-on. At the closest point, all kinetic energy converts to electric potential energy.
Given:
Solution:
At closest approach, kinetic energy equals electric potential energy:
Solving for r:
This is about 30 times larger than the actual gold nucleus radius (~7 fm), explaining why alpha particles can approach closely enough to experience strong repulsion.
In 1913, Niels Bohr developed a model of the hydrogen atom that successfully explained atomic spectra. While Rutherford's nuclear model was correct, it had a fatal flaw: classical physics predicted that orbiting electrons should continuously radiate energy and spiral into the nucleus within microseconds. Bohr resolved this by introducing quantum concepts.
Calculate: (a) the wavelength of light emitted when a hydrogen electron transitions from n = 3 to n = 2 (first line of Balmer series), (b) the ionization energy of hydrogen from the ground state.
(a) Balmer series: n = 3 → n = 2
Energy of initial state:
Energy of final state:
Energy of emitted photon:
Converting to Joules:
Using :
This is the red H-alpha line, close to the theoretical 656 nm
(b) Ionization energy from ground state:
Ground state energy:
Ionized state energy:
Ionization energy:
This is the minimum energy needed to remove an electron from ground-state hydrogen
In 1924, Louis de Broglie proposed a revolutionary idea: if light can behave as both wave and particle (as shown by the photoelectric effect), then perhaps matter particles can also exhibit wave behavior. This wave-particle duality extended quantum mechanics to all matter.
Calculate the de Broglie wavelength of: (a) an electron accelerated through 100 V, (b) a neutron at room temperature (thermal energy kT ≈ 0.025 eV). Compare with visible light wavelengths.
Given:
(a) Electron through 100 V:
Kinetic energy:
From , find momentum:
De Broglie wavelength:
Comparable to atomic spacing in crystals, perfect for diffraction experiments
(b) Thermal neutron:
Energy:
Both wavelengths (~0.1-0.2 nm) are much smaller than visible light (400-700 nm), which is why electron and neutron diffraction can probe atomic-scale structures that visible light cannot resolve.
In 1926, Erwin Schrödinger developed a comprehensive quantum mechanical description of the atom using wave mechanics. Rather than describing electrons as particles in specific orbits, the Schrödinger model uses wave functions to describe the probability of finding an electron at different locations.
For an electron in the n = 3 shell of hydrogen: (a) list all possible combinations of quantum numbers (n, ℓ, m_ℓ), (b) how many orbitals exist in this shell? (c) what is the maximum number of electrons this shell can hold?
(a) Quantum number combinations for n = 3:
ℓ = 0 (3s): m_ℓ = 0 → 1 orbital
ℓ = 1 (3p): m_ℓ = -1, 0, +1 → 3 orbitals
ℓ = 2 (3d): m_ℓ = -2, -1, 0, +1, +2 → 5 orbitals
Full list of (n, ℓ, m_ℓ) combinations:
(3,0,0), (3,1,-1), (3,1,0), (3,1,+1), (3,2,-2), (3,2,-1), (3,2,0), (3,2,+1), (3,2,+2)
(b) Total number of orbitals:
1 + 3 + 5 = 9 orbitals
General formula: n² orbitals in the nth shell
(c) Maximum electrons:
Each orbital can hold 2 electrons (spin-up and spin-down)
Maximum = 9 orbitals × 2 electrons/orbital = 18 electrons
General formula: 2n² electrons in the nth shell
In 1932, James Chadwick discovered the neutron, solving a major puzzle in nuclear physics. Before this discovery, the nucleus was thought to contain only protons and electrons, but this model couldn't explain nuclear masses or other observed properties.
The discovery resolved the atomic mass puzzle. For example, helium has atomic number 2 (two protons) but mass number 4. Previously, scientists thought the nucleus contained 4 protons and 2 electrons to give net charge +2, but this created theoretical problems. The neutron model explained it simply: helium has 2 protons and 2 neutrons, giving mass 4 and charge +2. The neutron became the third fundamental particle alongside the proton and electron.
A neutron (mass 1.675 × 10⁻²⁷ kg) moving at 3.0 × 10⁶ m/s collides head-on with a stationary proton (mass 1.673 × 10⁻²⁷ kg). Assuming an elastic collision, use conservation of momentum to estimate the final velocity of the proton. (For equal masses in elastic head-on collision, velocities are approximately exchanged.)
Given:
Analysis:
For elastic collision between particles of equal mass with one initially at rest, the moving particle stops and the stationary particle acquires the initial velocity of the moving particle.
Conservation of momentum:
Since :
Conservation of kinetic energy:
Solving both equations simultaneously gives:
The proton is ejected with approximately the initial neutron velocity, explaining Chadwick's observations of high-energy protons from paraffin.
Not all combinations of protons and neutrons form stable nuclei. The stability of a nucleus depends on the balance between the attractive strong nuclear force and the repulsive electromagnetic force between protons.
For each of the following unstable isotopes, predict the most likely decay mode and write the decay equation: (a) uranium-238 (Z=92, N=146), (b) carbon-14 (Z=6, N=8), (c) fluorine-18 (Z=9, N=9).
(a) Uranium-238 (Z=92, N=146):
Analysis: Very heavy nucleus (Z = 92, well above 82)
Prediction: Alpha decay (reduces both Z and N)
Decay equation:
Uranium-238 decays to thorium-234 by emitting an alpha particle
(b) Carbon-14 (Z=6, N=8):
Analysis: N/Z = 8/6 = 1.33 (above zone of stability for light nuclei where N ≈ Z)
Prediction: Beta-minus decay (converts neutron to proton, reduces N/Z ratio)
Decay equation:
Carbon-14 decays to nitrogen-14 (used in radiocarbon dating)
(c) Fluorine-18 (Z=9, N=9):
Analysis: N/Z = 9/9 = 1.0 (below zone of stability; needs more neutrons)
Prediction: Beta-plus decay or electron capture (converts proton to neutron, increases N/Z)
Decay equation (beta-plus):
Fluorine-18 decays to oxygen-18 (used in PET scans)
Radioactive decay is the spontaneous transformation of unstable nuclei into more stable configurations by emitting particles or electromagnetic radiation. The three main types of decay are alpha, beta, and gamma radiation.
Iodine-131 (used in medical treatments) has a half-life of 8 days. A hospital receives a 400 mg sample. Calculate: (a) the mass remaining after 24 days, (b) the time for the sample to decay to 25 mg.
Given:
(a) Mass after 24 days:
Number of half-lives:
After 3 half-lives, only 1/8 of the original sample remains
(b) Time to decay to 25 mg:
Since , we have
It takes 4 half-lives (32 days) for the sample to decay to 25 mg (1/16 of original)
When nucleons bind together to form a nucleus, the mass of the nucleus is slightly less than the sum of its individual nucleon masses. This "missing mass" has been converted to binding energy according to Einstein's mass-energy relation E = mc².
Calculate the binding energy and binding energy per nucleon for helium-4. Given: m_p = 1.00728 u, m_n = 1.00866 u, m(He-4) = 4.00260 u, 1 u = 931.5 MeV/c².
Given:
Helium-4 has Z = 2 protons, N = 2 neutrons
Step 1: Calculate mass defect
Total mass of separated nucleons:
Mass defect:
Step 2: Calculate binding energy
Step 3: Binding energy per nucleon
Helium-4 is particularly stable for a light nucleus, which is why alpha particles (helium nuclei) are commonly emitted in radioactive decay.
Nuclear fission and fusion are processes that release energy by rearranging nucleons into more stable configurations. Both processes convert mass into energy according to E = mc², but they work in opposite directions on the binding energy curve.
A nuclear power plant generates 1000 MW of power through U-235 fission (200 MeV per fission). Calculate: (a) the number of fission events per second, (b) the mass of U-235 consumed per day.
Given:
(a) Fission events per second:
Power = Energy per second, so:
Over 31 quintillion fissions every second!
(b) U-235 mass consumed per day:
Fissions per day:
Number of moles:
Mass (molar mass of U-235 = 235 g/mol):
Only about 1 kg per day due to the enormous energy density of nuclear reactions!
The Standard Model of particle physics is the theory describing the fundamental particles that make up matter and the forces through which they interact. It organizes particles into two categories: fermions (matter particles) and bosons (force carriers).
The Standard Model successfully explains electromagnetic, weak, and strong interactions. However, it doesn't include gravity, doesn't explain dark matter or dark energy, and doesn't explain why neutrinos have mass (originally thought to be massless). These limitations suggest the Standard Model is incomplete, and physicists continue searching for a more comprehensive theory.
Determine the quark composition and charge for: (a) a proton, (b) a neutron, (c) explain which force carriers are involved in beta-minus decay of a neutron.
(a) Proton composition:
Proton = uud (two up quarks, one down quark)
Charge calculation:
The proton has charge +1e as expected
(b) Neutron composition:
Neutron = udd (one up quark, two down quarks)
Charge calculation:
The neutron is electrically neutral
(c) Beta-minus decay forces:
Beta-minus decay:
At the quark level, one down quark converts to an up quark:
Force carrier: This process is mediated by the W⁻ boson (weak force carrier). The down quark emits a virtual W⁻ boson and becomes an up quark. The W⁻ boson then decays into an electron and electron antineutrino.
Beta decay demonstrates the weak nuclear force changing quark flavor
Particle accelerators are essential tools for probing the structure of matter at the smallest scales. They accelerate charged particles to extremely high energies before colliding them with targets or with each other, revealing fundamental particles and forces.
In a cyclotron, protons are accelerated using a magnetic field of 1.5 T. Calculate: (a) the cyclotron frequency, (b) the maximum kinetic energy if the dee radius is 0.5 m.
Given:
(a) Cyclotron frequency:
For circular motion in magnetic field:
This gives:
Period:
Frequency:
This frequency is independent of particle speed or radius!
(b) Maximum kinetic energy:
At maximum radius:
Kinetic energy:
Converting to eV: