A question about relaxation oscillator
Collector-emitter breakdown voltage isn't the relevant parameter in this circuit.

Note that the B-E junction is reverse-biased, and the B-C junction is forward biased, so the relevant parameter is the B-E breakdown voltage, which is just 6.0 V.

When the B-E junction breaks down in reverse bias, charge carriers are injected into the base region, which allows the transistor to operate in a "reverse active" mode, passing currrent from the capacitor through the LED. This continues until the capacitor voltage drops too low to sustain this.

This mechanism creates a negative-resistance characteristic in the I-V curve of the transistor in this mode of operation, and it is this negative resistance that creates the relaxation oscillator. Without it, the circuit would just come to equilibrium without oscillating.