= Solution
A <direct band gap> has its conduction minimum and valence maximum at the same <crystal momentum>. A photon can then connect the band-edge states with negligible momentum change. An <indirect band gap> requires an additional phonon to supply or remove momentum, making radiative recombination much less probable. Direct-gap materials such as <gallium arsenide> or <gallium nitride> are therefore usually better for a <light-emitting diode> than <silicon>.
For the graded junction, let $q=|e|>0$, $\varepsilon_s=\varepsilon_0\varepsilon_r$, and use electrostatic potential $\varphi$ with $E_x=-\varphi'$. Fully ionized dopants have charge density $\rho=q(N_D-N_A)=q\alpha x$, so <Poisson's equation> is $\varphi''=-q\alpha x/\varepsilon_s$. The paper's displayed plus sign corresponds to using the negative potential $V=-\varphi$, or a signed electron charge in that formula. The positive barrier magnitude and charge magnitude must be used consistently in the width below.
The intended compensation point is \b[$x=0$], where $N_D=N_A$ and the net doping changes sign. A locally neutral model with comparable electron and hole mobilities has its lowest carrier conductivity near there. The <depletion approximation> neglects mobile-carrier conductivity throughout the layer, so it does not by itself locate an exact conductivity minimum; unequal mobilities or asymmetric carrier profiles can shift it.
Let the depletion edges be $x=\pm d$. With zero exterior field, integrating $E_x'=\rho/\varepsilon_s$ and imposing $E_x(\pm d)=0$ gives the <linearly graded p-n junction> field
$$
\boxed{E_x(x)=\frac{q\alpha}{2\varepsilon_s}(x^2-d^2),\qquad |E_x|_{\max}=\frac{q\alpha d^2}{2\varepsilon_s}\text{ at }x=0.}
$$
Its integral gives $\varphi(x)=\varphi(0)+q\alpha(d^2x-x^3/3)/(2\varepsilon_s)$, so the <electric potential> increases toward the n-type side while the electron band edges $E_c,E_v$ bend downward together. Their separation is the local band gap, and the equilibrium <electrochemical potential> is flat. The <built-in potential> arises when carrier diffusion exposes ionized dopants until the opposing field balances further diffusion.
\Image[/past-exam-of-the-quantum-condensed-matter-course-of-the-university-of-cambridge/2007-graded-junction.png]
{title=Band bending and electric field in a symmetric linearly graded junction}
{description=Original schematic band edges and a flat equilibrium chemical potential. The lower panel plots the normalized parabolic depletion field; outside the edges it is zero. Energy offsets are illustrative, while the cubic potential and field shape follow from the graded-charge model.}
Finally the positive barrier is
$$
V_b=\varphi(d)-\varphi(-d)=-\int_{-d}^dE_x\,dx=\frac{2q\alpha d^3}{3\varepsilon_s}.
$$
With total width $w=2d$, this yields the <linearly graded junction depletion width>
$$
\boxed{w=\left(\frac{12\varepsilon_0\varepsilon_rV_b}{q\alpha}\right)^{1/3}.}
$$
Here $\alpha>0$ labels p-to-n progression with increasing $x$; otherwise use its magnitude in the width formula.
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