The radio frequency source can be modeled as a constant voltage source with a series characteristic impedance of say 50 Ohms. That means it can drive a 50 Ohm load.
Depends on the radio frequency source.
A signal generator which has a maximum output of 2 or 3 volts can be modeled as you suggest, it can drive any impedance load it likes.
A power amplifier could be designed to work into a 50ohm load but that does not mean that the output impedance is 50ohm.
Q1.Does that mean the E/H ratio is 50 as in vacuum E/H is 120π.
I have never heard of E/H ratio.
A quick googling for E/H ratio gave references to photoelectrics and hadron calorimeters, you seem to be making some incorrect comparison with another branch of physics.
Secondly, when source is perfectly matched to a reactive load say 1+j45 ohms by a matching network, by maximum power transfer theorem, half of the power is transferred. Q2.Then where the other half power is dissipated?
If the generator
CAN be accurately modeled as a 50 ohm resistor in series with a perfect voltage source, then the other half of the power is dissipated in the resistor.
Dont lose sight of the fact that this is just a theoretical generator, a practical generator will be more complex.
I have used an L section matching at frequency of 13.56 MHz.
My detector says 100% forward power and 0% reflected power for an input RF power of 1 kilowatt.
So, you are using a power amplifier (1kW) on one of the ISM frequencies, I would suggest that the amplifier requires a 50ohm load, rather than it has a 50ohm output impedance.
Your "detector", I assume is some kind of VSWR meter or a directional power meter and shows that your matching network is doing a good job of converting your complex load to 50ohm resistive, which is what the generator requires to work correctly.
The efficiency of the generator is another question entirely.
Comparing power input from the mains outlet on the wall to RF out from the output connector the efficiency will be quite a lot less than 50%.
JimB