AM radio wave LED indicator
Since you are using an NPN transistor, you should have it between the LED and the negative pole of the battery instead of the positive pole.
Like this:



simulate this circuit &ndash; Schematic created using CircuitLab
An NPN transistor requires the base to be about 0.7 V above the emitter voltage so that it will conduct.
To get that in your circuit, you have to get Vin up to 0.7V (for the base voltage) plus the forward voltage of the LED (somewhere around 1.5V.)
To make your LED light up, you have to get about 2.2V on the base of the transistor.  Your detected radio signal has to produce 2.2V - that takes a good antenna and a strong transmitter close by.
The second circuit reduces that requirement somewhat.  It only needs 0.7 V at Vin to make the transistor conduct.
Still, 0.7V is a lot.
What you really need is a way for the transistor to work with much smaller voltages.
The way to do that is just like they used to do it in single transistor AM radio receivers:

To make it light your LED, you would replace the earphone with your LED.
Like this:



simulate this circuit
The values are just eyeballed - you'll have to see what works.

Another thing that can be a problem is the simulator.
Depending on what mathematical model of the diode it uses, it may or may not actually be able to simulate rectification of signals smaller than about 0.7V.
The old germanium diodes would rectify voltage at much lower than the typically quoted 0.3V forward voltage - if the current was very low.
Forward voltage depends on the current - lower current, lower forward voltage.
This earlier question goes into some detail about diodes and forward voltage.
If your simulator's model doesn't accurately simulate the low current behavior of your diode, then the simulation will show no output for a circuit that would work in real life.