Will this transistor work ?

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But you must understand that normally we put the relays between collector and emitter

**broken link removed**

And is this case the protection diode will do the work. The diode will protect the BJT.

But here we have a very different situation.

1. The relay is connect between BJT emitter and the GND.

2. Also we have a fairly low resistor connect between the BJT base and the GND.

So to cut-off the BJT LDR resistor must be illuminated by a strong beam of light.
Before this happens, the current through relay coil will flow from point A to point B.
Once the transistor is at cut-off, the relay coil tries to keep the current to flow in the same direction as before. This means that the voltage at point B goes positive relative to voltage at point A
And when this voltage reach 0.7V the BJT goes into conduction mode again.
The BJT will be ON until relay coil "discharge" their energy.


Of course in real life the induce voltage will be higher then 0.7V. Simply because I ignored Rb (LDR) voltage drop.
 
The reasoning above assumes the transistor is switched off rapidly. However when the LDR changes resistance, the change is so slow that the back emf from the relay-coil is never more than the applied voltage and no spikes are produced.
 
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