Redshift Academy

Wolfram Alpha:         

  Search by keyword:  

Astronomy

-
-
-
-

Chemistry

-
-
-
-

Classical Mechanics

-

Classical Physics

-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-

Climate Change

-

Cosmology

-
-
-
-
-
-
-
-
-
-
-
-
-
-

Finance and Accounting

-
-
-
-
-
-
-
-
-

Game Theory

-

General Relativity

-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-

Group Theory

-
-
-
-
-
-

Lagrangian and Hamiltonian Mechanics

-
-
-
-
-
-

Macroeconomics

-
-
-

Mathematics

-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-

Mathjax

-

Microeconomics

-

Nuclear Physics

-
-

Particle Physics

-
-
-
-
-
-
-

Probability and Statistics

-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-

Programming and Computer Science

-
-
-
-
-
-

Quantitative Methods for Business

-

Quantum Computing

-
-
-

Quantum Field Theory

-
-
-
-
-

Quantum Mechanics

-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-

Semiconductor Reliability

-

Solid State Electronics

-
-
-
-
-

Special Relativity

-
-
-
-
-
-
-
-
-
-
-
-

Statistical Mechanics

-
-
-

String Theory

-
-
-
-
-
-

Superconductivity

-
-
-
-
-
-

Supersymmetry (SUSY) and Grand Unified Theory (GUT)

-
-
-
-
-

The Standard Model

-
-
-
-
-
-
-
-
-
-

Topology

-

Units, Constants and Useful Formulas

-
Last modified: January 26, 2018

Nuclear Spin ------------ By analogy with the Bohr magneton we can define the NUCLEAR MAGNETON as: μN = eh/2mp For individual electrons, protons and neutrons we can write: μe = gμBS/h where g = -2.0023 = gμBms μp = gμNS/h where g = 5.585691 = gμNms μn = gμNS/h where g = -3.826084 = gμNms Note the neutron has 0 charge but has a magnetic moment because the component quarks are charged. For combinations of neutrons and protons into nuclei, the situation is more complex. There is no simple formula to predict the nuclear spin based on the number of protons and neutrons within an atom. Nevertheless, there are some general rules that apply to special cases: # of protons # of neutrons Spin Examples ------------ ------------- ---- -------- Even Even 0 12C, 16O Even Odd m/2 17O Odd Even m/2 1H, 23Na, 31P Odd Odd m 2H Where m is an integer. If these magnetic moments are placed in a static magnetic field they will have a potential energy related to their orientation with respect to that field. In the case of an electron, the lowest energy configuration is when the dipole moment is aligned with the field. The maximum energy is when the dipole is in anti-alignment. Due to the difference in charge, the opposite is true for the proton. In both cases, there will be precession around the mgnetic field. The Larmor frequency is given by: ωp = μpB/h Quantum mechanically the precession can be thought of as the quantum energy of transition between the two possible spin states for spin 1/2. The process of transition can be expressed as a photon energy according to the Planck relationship. Thus, E = hωp = μpB ------- +μpB ΔE = 2μpB ------- -μpB The energy difference between the 2 spins states is very small in comparison to the thermal energy at room temp. This means that the degree of polarization that can be maintained at ordinary temperatures is extremely small. Nuclear Magnetic Resonance -------------------------- The precession of the proton spin is used NMR. Hydrogen is placed in a strong magenetic field to polarize the proton spins and then an RF field is applied to excite some of the nuclear spins into a higher energy state. The absorbtion of the RF energy only occurs when the frequency of the radiation equals the Larmor frequency - hence the term 'resonance'. When the RF field is removed, the spins tend to return to their lower energy state (spin relaxation) in the process emitting a small amount of radiation at the Larmor frequency.