Um.... I'll try the radio shack thing first.
I have a friend who's an EE. here is what he told me......
Simple explanation: Yes, but you will need a resistor in series of about 560 Ohms.
Technical Details: An LED (or any diode) will conduct infinite current (0 Ohms resistance) with a a fairly fixed voltage drop (volts) when voltage is applied. Recall a couple of formulas:
1) Power = Volts * Amps
2) Volts = Amps * Ohms
Given that an LED has a resistance of 0 will mean (theoretically) a current (Amps) of Infinity. This will mean that the power in equation one will go to infinity and burn something out (either a fuse or the LED -- probably the LED).
I noticed that this LED has a forward drop (voltage drop across it) of 3.17Volts and a current of .020 Amps from some of the comments from users on the webpage. How can I create this happy operating state? Lets examine equation #2:
Volts = Amps * Ohms
I know that Volts is fixed ~14.4 in a car and I know that when I put .020 Amps into the diode it will drop 3.17 volts and I cannot change these numbers. The volts side is now:
(14.4 - 3.17) = Amps * Ohms
I know the LED likes .020 Amps, so now I have:
(14.4 - 3.17) = .020 * Ohms
Ohms = (14.4-3.17)/.020 = 561
The standard resistor sizes for 5% accurate resistors are 510, 560, 620, and 680. I would try a 560. If it is too bright, go to a 620. If it is not bright enough, go to a 510. A smaller resistor will allow more current and thus a brighter light although at some point you will begin to reduce the longevity of the LED. Of course, like anything in engineering all of the equations above shift over temperature.
Here is a nice page on the topic:
(you would use 14.4, 3.17, and .020 in this simulator)
PS. It is important which end of the LED is connected to + and - voltage, but it does not matter if the resistor is between the battery and the LED or the LED and ground. Have fun