The two-transistor model describes a thyristor, and the condition that marks the boundary of latching is α₁ + α₂ = 1 — option 3.

The model. A four-layer pnpn structure is split into an interleaved pnp and npn pair, each transistor's collector feeding the other's base. Writing the anode current in terms of both alphas and the leakage gives

\(I_{A}=\dfrac{\alpha_{2}I_{G}+I_{CBO1}+I_{CBO2}}{1-\left(\alpha_{1}+\alpha_{2}\right)}\)

Everything follows from the denominator. As \(\alpha_{1}+\alpha_{2}\) approaches 1 the denominator approaches zero and the anode current rises without bound — that is, it becomes limited only by the external circuit, which is exactly what "turning on" means. So

\(\alpha_{1}+\alpha_{2}=1\)

is the trigger condition, the precise threshold at which regeneration takes over.

Why option 1 is a trap worth examining. The regenerative loop needs a loop gain of at least 1 to start, so one might argue for \(\gt1\). But the sum cannot stay above 1: once conduction begins, the large current drives both transistors into saturation, where the alphas fall back and the sum settles at exactly unity. The equality is both the trigger condition and the steady conducting condition, which is why it is the answer.

Why the alphas rise in the first place. Current gain in a bipolar transistor is small at very low currents, because recombination in the emitter-base depletion region takes a large share of the current. As current increases, that share falls and \(\alpha\) climbs towards its full value. This is the mechanism behind every way of firing a thyristor:

Gate current raises \(I_{G}\) and hence the npn section's current directly.
Break-over voltage multiplies the leakage by avalanche.
High dv/dt injects displacement current \(C_{j}\,dv/dt\).
Temperature raises the leakage.

All four do the same thing — push the alphas up until their sum reaches unity.

Why turn-off is hard follows from the same equation: nothing in it involves the gate once conduction has begun, so the anode current must be reduced below the holding value to bring the sum back under 1.

Hence, α₁ + α₂ = 1.