Pedantically, assuming we are driving each of these 1 watt LEDs to their full power handling capability, they put out exactly the same radiant flux: 1 watt, or 1 joule per second. Then again, so does a 1W resistor.
The reason: with 1W of electrical power going in, there must be at equilibrium 1W of power going out. Some of it will be red light, and the rest of that electrical power will go into heating the component until blackbody radiation is enough to achieve thermal equilibrium. Within the normal operating temperatures of these components, that blackbody radiation will be in the far infrared spectrum, but the definition of radiant flux includes all electromagnetic radiation, even that which is not visible.
The LEDs will emit less photons than the resistor, because at least some of their radiation will be at a higher frequency. Higher frequency photons have more energy, so fewer are required.
But probably, you care just about the visible radiation from the LEDs. To do this, you have to define just what you mean by "visible". A standard definition is luminous flux:
In photometry, luminous flux or luminous power is the measure of the perceived power of light. It differs from radiant flux, the measure of the total power of electromagnetic radiation (including infrared, ultraviolet, and visible light), in that luminous flux is adjusted to reflect the varying sensitivity of the human eye to different wavelengths of light.
For sure, either of your 1W LEDs will have a much higher luminous flux than a resistor. The 630nm LED has a bit of an edge over the 660 nm LED as the eye is more sensitive to 630nm (peak sensitivity being somewhere around 550 nm). However, this is but one of many variables needed to answer your question. The particulars come down to the specifics of the device. Read the datasheets.
To address some of your additional concerns:
I'm very confused with the following Higher wavelength == less energy, right?
Less energy per photon, but this is really irrelevant for everyday purposes.
$$\text{energy per photon} = h\nu = \frac{hc}{\lambda} $$
Where:
- \$c\$ is the speed of light in meters per second
- \$h\$ is the Planck constant
- \$\nu\$ is the frequency in hertz
- \$\lambda\$ is the wavelength in meters
See Energy vs. power in transmitters from Amateur Radio Stack Exchange. Although that's about radio transmitters, it's still electromagnetic radiation just like visible light (but at a much lower frequency)
Higher wavelength also == less number of photons coming out, right?
Assuming for this comparison an equal radiant flux. You've got it backwards: higher wavelength means a lower frequency, thus less energy per photon. To attain a given radiant flux (power), you can have:
- a higher number of lower energy (higher wavelength, lower frequency) photons, or
- a smaller number of higher energy (lower wavelength, higher frequency) photons
being emitted each second. The number of photons emitted per second, multiplied by the energy per photon equals the radiant flux:
$$ \text{photons per second} \cdot \text{energy per photon} = \text{radiant flux} $$
I am concerned with output light in the visible spectrum, from around 400 to 700nm range. However, I am not concerned with the sensitivity of the human eye.
If this is the case, and you want to weight all radiation in this visible range equally, you can do that. In this case, there's really very little difference besides color between a 660nm and 630nm LED (or an LED of any other color). Any differences in efficiency by this definition will be a consequence of the device's construction. For example, the resistance of the leads, the shadowing of the LED die by the bond wires, the opacity of the lens, flaws in the manufacture, etc.
If it takes longer to create a single photon of 660nm, which has less energy than 630nm does the energy difference between 630nm and 660nm photon is turned into heat by LED?
No. Although a 630nm photon and a 660nm photon have different energies, the rate they are emitted will be different such that the radiant flux will be identical.
Think of it is this way: if you have a factory that consumes raw water at 100 liters per second and has to ship that water out at the same rate, you can:
- ship 100, 1-litre bottles per second
- ship 10, 10-litle bottles per second
- or any number of other combinations, as long as your total water output rate is equal to 100 liters per second.
Each photon is a package of energy. The LED consumes electrical energy at some rate (1W), and it must output some other kind of energy at the same rate (probably emitting photons, either as visible light or blackbody radiation). It doesn't matter what size the packages are.
