Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2017 / S16

A weak positive potential perturbs Boltzmann electrons as $n_e\simeq n_0[1+e\varphi /(k_{B}T)]$ against nearly fixed ions. Poisson's equation outside the test charge becomes ${\nabla }^2\varphi =\varphi /{\lambda }_D^2$, withThe resulting potential is proportional to $e^{-r/{\lambda }_D}/r$. The Debye length is the screening distance over which charge imbalance is suppressed, within this classical weak-perturbation model.

Define the plasma parameter as the number in a Debye sphere, $N_D=(4\pi /3)n_e{\lambda }_D^3$; some conventions omit $4\pi /3$. A good classical plasma has many particles in a screening volume, $N_D\gg 1$, and a system size much larger than ${\lambda }_D$, so collective fields dominate individual encounters.

A collision here means cumulative Coulomb deflection that significantly changes a particle's velocity, not necessarily a literal hard-sphere impact. A collisionless description requires the phenomenon's timescale to be short compared with the appropriate collision time, and its length scale short compared with the mean free path. Parametrically $\boxed{{\omega }_p{t}_c\sim N_D/\log \Lambda }$ up to convention-dependent constants, with Coulomb logarithm $\log \Lambda$ representing the range of impact parameters. Many oscillations can then occur before collisional relaxation. The formula itself presupposes the weak-coupling regime.

Rapid longitudinal oscillations in a weakly collisional plasma compress the velocity distribution primarily along one direction. There is insufficient time for collisions to redistribute that energy among three directions; the effective one-dimensional adiabatic exponent is $1+2/1=3$. Ordinary monatomic-gas sound with local collisional equilibrium has three translational degrees of freedom, giving $1+2/3=5/3$. Thus $\Gamma =3$ belongs to the stated kinetic/time-scale limit, not to every thermodynamic process involving ionized gas.

The ion density is $\rho /(9m_u)$ and double ionization gives $n_e\simeq 2.48\times 10^{29}$ m${}^{-3}$. Therefore $\hslash {\omega }_p\simeq 18.5$ eV. The photon energy loss gives $\hslash \omega =30$ eV, while momentum conservation givesInverting the stipulated dispersion law givesAn angular frequency expressed “in eV” means $\hslash \omega$, not the frequency itself. The temperature is a formal inference conditional on the dispersion model, whose validity is tested next.

The inferred parameters give ${\lambda }_D\simeq 6.46\times 10^{-11}$ m and $\boxed{N_D\simeq 0.28}$, or $n_e{\lambda }_D^3\simeq 0.067$ without the spherical factor. Neither is much larger than one. Thus a good classical weakly coupled plasma approximation is not justified. Correlations, collisions and potentially degeneracy matter at this density; treating the preceding 18.7 eV as a reliable temperature measurement would overstate what this simple model establishes.

At half ionization, electron and neutral densities are each $n_0/2$, so the left side of Saha's equation is $n_0/2$. Its logarithm changes by $d\log n_0=(3/2+R/(k_B{T}^{\prime }))d\log T^{\prime }$. To first order,This holds for small fractional changes and the supplied ground-state partition approximation, not an exact finite-difference identity.

The measured ratio is ten, so $R/(k_B{T}^{\prime })=8.5$. Taking $R=13.6$ eV gives $k_B{T}^{\prime }=1.60$ eV or $T^{\prime }\simeq 1.86\times 10^4$ K. Substitute into Saha's equation:Higher density favors recombination and reduces the entropy advantage of free electrons at a fixed ionized fraction, so a higher temperature is required. The ratio 8.5 means half ionization does not require $k_{B}T$ to equal the ionization energy: translational phase space and the exponential tail are important.

Hydrogen's $n=2$ excitation energy is $3R/4=10.2$ eV. With electronic degeneracy ratio $g_2/g_1=4$, its relative population is $4e^{-10.2/1.60}\simeq 6.8\times 10^{-3}$. Thus the first excited manifold contributes below one percent of the ground population at the inferred temperature, explaining the leading ground-state approximation.

There is a qualification to extending this argument to infinitely many Rydberg states: the isolated-atom partition sum with degeneracy $2n^2$ formally diverges. Plasma screening and level dissolution impose a physical cutoff. The small low-lying populations justify the simple approximation in its intended regime, not an unrestricted infinite partition sum.

Use a steady planar sonic ablation model with absorbed intensity $I_a$, critical mass density ${\rho }_c=(A_i/Z)m_u{n}_c$, and $n_c={ϵ}_0{m}_e(2\pi c/\lambda {)}^2/e^2$. In the usual isothermal-electron sonic balance, $I_a\simeq 4{\rho }_c{c}_s^3$ and the ablation pressure is $P_A\simeq 2{\rho }_c{c}_s^2$. Eliminating $c_s$ givesFor $A_i/Z\simeq 2$ and complete absorption, the supplied laser parameters give about $1.15\times 10^{13}$ Pa, or $115$ Mbar. Mach one alone fixes neither the material mass per electron nor the enthalpy/absorption coefficient. Different sonic-flow closures change the order-one constant; incomplete absorption multiplies the estimate by its fraction to power $2/3$. The robust result is the two-thirds scaling, rather than a material-independent numerical pressure.

An inertially confined sphere disassembles on approximately $\tau \sim r/c_s$, while $n_i=\rho /{\bar{m}}_i$. Substitute these in the given number-density confinement product to getFor an equal D-T mix, ${\bar{m}}_i\simeq 2.5m_u$. Taking $c_s\sim \sqrt{2k_{B}T/{\bar{m}}_i}$ at 10 keV gives $A\simeq 0.365$ kg m${}^{-2}=0.0365$ g cm${}^{-2}$. An adiabatic sound-speed convention changes this by an order-one factor. This estimate implements the supplied Lawson product and is not the separate, more stringent alpha-stopping areal-density criterion.

The final radius is $r_f=0.5\mathrm{mm}/15=33.3\mu$m. For a 3.5 MeV alpha particle, $p\simeq \sqrt{2m_{\alpha }{E}_{\alpha }}$ and the largest perpendicular gyroradius is $p/(2e{B}_f)$. Setting it equal to the core radius givesThis is an order-of-magnitude onset estimate for magnetic deflection, assuming flux freezing and a suitably oriented field. A small gyroradius alone does not ensure complete confinement along open field lines or efficient deposition for every pitch angle.

Magnetic pressure is the field energy density $P_B=B_f^2/(2{\mu }_0)$. Using the preceding final field gives $\boxed{P_B\simeq 2.6\times 10^{13}\mathrm{Pa}\simeq 260\mathrm{Mbar}}$. This uses the compressed field, not the initial 36 T field; substituting the latter would omit the square of the already squared convergence amplification.