What "degeneracy" means here. A tunnel diode is made by doping both sides so heavily (10¹⁹–10²⁰ cm⁻³) that the Fermi level moves into the bands — into the conduction band on the n-side and into the valence band on the p-side. The degeneracy is the amount by which the Fermi level penetrates, measured in volts:

\(V_n=\dfrac{E_F-E_C}{q}\)

Why the simple Boltzmann formula is not enough. For a lightly doped (non-degenerate) semiconductor,

\(E_F-E_C=kT\ln\!\left(\dfrac{N_D}{N_C}\right)\)

This is negative while \(N_D \lt N_C\) (the Fermi level lies below the band edge) and only turns positive once the doping approaches the effective density of states N_C. But at such high doping the carriers obey Fermi–Dirac rather than Boltzmann statistics, so a correction term is added:

\(V_n=\dfrac{kT}{q}\left[\ln\!\left(\dfrac{N_D}{N_C}\right)+0.35\left(\dfrac{N_D}{N_C}\right)\right]\)

The constant 0.35 is the Joyce–Dixon coefficient \(1/\sqrt{8}=0.3536\), the leading term of the series that corrects the Boltzmann approximation as degeneracy sets in.

How to pick the right option without recalling the constant. Two structural checks do it:

The ratio must be N_D/N_C. Degeneracy grows with donor concentration, so N_D belongs on top — that eliminates options 3 and 4.

The correction must be additive. Fermi–Dirac statistics place the Fermi level higher than Boltzmann predicts at high doping, so the correction adds — eliminating option 2.

Why this matters for the device. Both sides being degenerate is what puts filled states on one side directly opposite empty states at the same energy on the other, across a depletion layer only ~10 nm wide. That is what permits tunnelling at low forward bias and produces the negative-resistance region used in high-frequency oscillators.

Hence, the degeneracy on the n-side is \(\dfrac{kT}{q}\left[\ln\left(\dfrac{N_D}{N_C}\right)+0.35\left(\dfrac{N_D}{N_C}\right)\right]\).