Past exam of the particle physics course of the University of Oxford / 2018 / 5

A fast charged particle loses energy by ionizing and exciting silicon, creating electron–hole pairs along its path. In a reverse-biased microstrip diode, the electric field separates and drifts the electrons and holes toward opposite electrodes. Their motion induces a current on the strips, which is integrated by the readout electronics to locate the hit.

The bias depletes the bulk of mobile equilibrium carriers, reducing conduction and capacitance, and supplies rapid directed charge collection before recombination or trapping. Without a collecting field, diffusion is slow and charge spreads or recombines, degrading signal and position measurement. The signal is generated by the traversing particle's energy deposition, not by the bias creating the ionization charge.

Treat the $p^{+}$ strips as much more strongly doped than the $n$ bulk, so the depletion layer lies predominantly in the latter. Fully ionized donors leave uniform space charge $\rho =e{N}_D$. In one dimension,At the depletion boundary $x=d$, the minimum-bias solution has $E(d)=0$. Thus $E(x)=e{N}_D(x-d)/{ϵ}_{\mathrm{Si}}$, and the magnitude of the potential drop isFor $N_D=10^{18}{\mathrm{m}}^{-3}$, $d=300\mu \mathrm{m}$, and ${ϵ}_r=11.6$, this isThe applied reverse bias is approximately this value; more precisely the built-in junction potential contributes to the total depletion voltage and is subtracted from the required external bias. It is small on the scale of seventy volts and is not specified. Here the stated free-carrier density is used as the effective donor density of the neutral bulk before depletion.

For binary strip readout, report the centre of the hit strip. With uniform impact position across pitch $w$, the position error is uniformly distributed on $[-w/2,w/2]$, soThis assumes no analogue charge sharing between strips.

The magnetic force is $d\mathbf{p}/dt=q\mathbf{v}\times \mathbf{B}$. Since relativistic momentum is parallel to velocity, $d(p^2)/dt=2\mathbf{p}\cdot q(\mathbf{v}\times \mathbf{B})=0$. A uniform transverse field therefore changes momentum direction but not magnitude. If the trajectory turns at angular speed $v/R$, then $|d\mathbf{p}/dt|=p_{T}v/R$. Equating force magnitudes yieldsThis is the positive radius. The sign of $q$ determines the direction of curvature; the source's signed formula can describe an oriented radius. In practical tracking units,

The source labels $L$ as the full chord from $P_1$ to $P_3$, with $L/2$ on each side of its midpoint. The Right triangle between the Circle centre, chord midpoint and endpoint givesFor fixed geometry and field, the inverse transverse momentum isIf the uncertainty is dominated by the central point, ${\sigma }_s\simeq {\sigma }_x$. Linear propagation of uncertainty givesUsing $|q/e|=1$, $B=2\mathrm{T}$, $L=0.6\mathrm{m}$ and ${\sigma }_x=28.87\mu \mathrm{m}$,The inverse-momentum resolution is approximately momentum independent in this spatial-error-only model; the relative momentum resolution grows as $p_T\sigma (1/p_T)$. Endpoint errors and multiple scattering are excluded by the stated assumptions.