= Solution
Take a <p-n junction> made from one <semiconductor>, with the <p-type semiconductor> on the left and the <n-type semiconductor> on the right. At <thermal equilibrium> the <chemical potential> $\mu=E_F$ is horizontal throughout. Before contact the <chemical potential> lies nearer the <valence band> on the p side and nearer the <conduction band> on the n side; after contact the band edges bend to permit one common value.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2022/paper-2-a4-bands.png]
{title=Original equilibrium <p-n junction> <energy band> diagram, <impurity levels>, common <chemical potential> and <depletion region>}
In this orientation both the <conduction band> edge $E_C$ and the <valence band> edge $E_V$ decrease from p to n by $eV_{\rm bi}$. For a <homojunction> their separation $E_g=E_C-E_V$ is constant. The <acceptor impurity> level $E_A$ lies just above $E_V$ where acceptors are present, and the <donor impurity> level $E_D$ lies just below $E_C$ where donors are present. These <impurity levels> follow the local <band bending>; they do not form one additional horizontal band across the device.
The shaded <depletion region> has few mobile <Electrons> or <electron holes> and contains fixed negative ionized <acceptor impurities> on the p side and positive ionized <donor impurities> on the n side. Thus its <electric field> points from n to p. Its neutral outer regions have negligible <electric field>, so the band edges become flat there. In the <depletion approximation>, <charge neutrality> requires $N_Ax_p=N_Dx_n$; the symmetric drawing illustrates $N_A=N_D$ rather than assuming that equality is universal. <Diffusion> and field-driven <drift current> cancel at equilibrium: no applied <voltage> does not mean zero internal <electric field> or zero built-in <electric potential difference>.
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