Extraction energy is 6000kg×2500kWhkg−1=15 GWh per year. A continuously operating 2 GW electrical plant generates 17.5 TWh per year, so lithium extraction consumes about 0.086% of its output under the stated estimate. This is energetically favorable even before claiming anything about capital costs or extraction feasibility.
For an ocean mass about 1.4×1021 kg, concentration0.17 ppm implies about 2.4×1014 kg lithium. At the specified consumption/output ratio, a hypothetical 20 TW electric world would use 6×107 kg annually, giving roughly 4×106years. For 3 TW the corresponding scale is 3×107years. State the assumed demand: “power the world” does not uniquely specify electric demand versus primary-energy use. These are inventory quotients ignoring growth, accessible fraction, ecological constraints and processing throughput.
Lithium in the blanket breeds tritium through neutron reactions, notably 6Li+n→4He+T. Tritium is not abundant enough in nature to supply a large reactor fleet without breeding. The blanket must also transfer heat and manage neutron damage and losses, and the breeding ratio must exceed consumption after accounting for processing and inventory. Lithium is therefore a fuel-cycle resource rather than the primary charged plasmareactant in the D-T core.
Essential systems include the vacuum vessel, toroidal and poloidal magnets, current-drive/transformer and auxiliary heating equipment, fueling and exhaust systems, divertor for concentrated heat and particle removal, neutron-absorbing/breeding blanket and shielding. Alpha particles retained in the plasma contribute self-heating. Confinement and controlled exhaust are as important as reaching a high temperature; a magnetic field does not confine the neutrons or make material heat loads vanish.
The radius is the electron-nucleus separation. For positronium, Z=1 and μ=me/2, so rn=2a0n2 and En=−Ry/(2n2). Each constituent's radius about the center of mass is half the separation; calling both that radius and rn the orbital radius without distinction would create another factor-two error.
A heated cathode emits electrons into low-pressure atomic vapor, and a variable accelerating voltage gives them kinetic energy. A grid and small retarding region precede the collectinganode. Below the first excitation threshold most collisions are elastic; above it, electrons can transfer a definite excitation energy to atoms and may then fail to overcome the collector's retarding field. Collector current develops repeated minima as accelerating voltage increases, separated by roughly the same excitation voltage, approximately 4.9V for mercury. Contact potentials and energy spreads shift or smear the minima without removing their characteristic spacing. A smooth loss mechanism would not produce these repeated fixed-energy thresholds.[ref]
Time evolution gives e−iE0t/ℏ[ϕ0+e−iωtϕ1]/2. It becomes the opposite relative-sign superposition when e−iωt=−1, first at tmin=π/ω. The overlap with the initial state is proportional to 1+e−iωt and vanishes then; global phase is immaterial.
Both probabilitydensities are even, so ⟨x⟩=0. The supplied Gaussian moments give ⟨x2⟩0=σ2/2 and ⟨x2⟩1=3σ2/2. Therefore Δx0=σ/2,Δx1=3/2σ. The original plots mark these two symmetric rms intervals around the common mean.
Squaring the states gives p0=e−x2/σ2/(πσ),p1=2x2e−x2/σ2/(πσ3). The first peaks at zero; the secondhas a node there and peaks at x=±σ. For a centered interval of width δx≪σ, P0≃δx/(πσ), whereas P1≃(δx)3/(6πσ3). Its leading linear term vanishes, but the probability in a finite interval is not exactly zero.
For the Gaussian, ϕ0′′=(x2/σ4−1/σ2)ϕ0. Using ℏ/(mω)=σ2, the x2 terms cancel and leave E0=ℏω/2. For ϕ1∝xϕ0, differentiation gives ϕ1′′=(x2/σ4−3/σ2)ϕ1, and the same cancellation gives E1=3ℏω/2. The normalization constants do not affect these eigenvalue equations.
Integrating F=−kx gives V=kx2/2up toa constant. Thus −ℏ2ϕ′′/(2m)+mω2x2ϕ/2=Eϕ, with ω=k/m. The potential is a symmetric upward parabola and tends to infinity in both directions; normalizable states decay at large ∣x∣.
To distinguish 2b equal transverse bins across width w, d=w/2b. Six bits gives one-nanometre pixels; ten gives 0.0625 nm, below the heliumwavelength. In the stipulated fixed-energy freely propagating beam, the latter entails transverse diffraction momenta comparable to or exceeding the available total momentum and lies outside the paraxial aperturemodel. Six-bit localization is less demanding, although fine detector resolution and losses remain practical constraints. This is not a universal theorem that an atom can carry at most six bits in every encoding: accessible modes, energy, aperture, measurements and success probabilitiesdefine the channel.
For an aligned narrower slit placed immediately after the first, uniform illumination gives transmission probabilityd/w. Conditional on passage the same normalized formula applies with w replaced by d and central-lobe momentum scale h/d. Position rms is d/12, while the exact hard-edge rms Δpx is infinite. Thus the source's finite “uncertainty” can only meana characteristic diffraction width; the rigorous Heisenberg inequality is not violated, and smoothing the edges restores finite rms moments satisfying it.
Immediately after a uniformly illuminated hard slit, the normalized transmitted state is ψ(x)=w−1/2 for ∣x∣<w/2, zero elsewhere. With Fourier convention ψ(p)=(2πℏ)−1/2∫e−ipx/ℏψ(x)dx,
The supplied sinc integral verifies unit normalization. The first zeros occur at ±δpx, but this finite central-lobe width is not the rms uncertainty: px2p fails to have a convergent integral.
Approximate heliummass by 2mp+2mn≃6.69×10−27 kg. Nonrelativistic momentum is p=2mT, giving λ=h/p≃0.203nm. A transverse width of one micrometre implies a characteristic position uncertainty of that order and a minimum momentumuncertainty of order ℏ/a∼10−28 kg ms−1, with factors depending on profile convention. For a uniform interval Δx=a/12, but an exactly abrupt rectangular wavefunction actually has infinite rms momentum spread. Width alone does not uniquely specify both rms uncertainties; a smooth near-minimum-uncertainty beam is needed for a finite order-of-magnitude estimate.