Is it possible to measure the resistance of n-Si or p-Si using multimeter?

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As the title says, is it possible? I was thinking that in order for the multimeter to measure resistance, it first provides current and checks the voltage, and then by ohm's law gets the resistance.

But would it even be possible to provide current to n-type Si bar or p-type Si bar? I know that in the diode, when the positive and negative leads are connected to p-type and n-type respectively, you can provide current. But what's going to happen if both positive and negative are connected to just n-type or just p-type?
 
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Yes, you can do that.

I tested a piece of Silicon which was supposed to be just elemental Silicon and should have been an insulator.

However, it was quite a good conductor.

Presumably it had been sourced from a semiconductor supplier and had deliberate impurities in it. I could not tell if it was N-type or P-type.
 
Yes a multimeter will do, but the result will be inaccurate. The voltage drop across the lead will also be included in measurement. Better use Kelvin's four probe method with corrections for probe placements, for much accurate result.

Make sure you solder the probes with Si to get good contact. Usually metal pressed against Si will have some small discontinuity at the interface sufficient to introduce high resistance. Also check whether the probe material produces Schottky contact with n and p silicon.

You can connect the diode in reverse and still get a current. only thing is that the current will be tiny. Well its a property of the diode. But your material is purely resistive. Thus you can connect any way you chose.
 
@ vk6kro - If you have a Si wafer it will be almost always evenly p type doped. It won't be intrinsic, or semi-insulating unless you're using GaAs wafer. If you have Si ingot it will be non evenly doped. The ingots and wafers are doped deliberately. However if you have MGS piece of Si it will have many unwanted impurity, defects and stuffs.