Today, we largely use LED lighting in our homes and streets. However, in the not-too-distant past, the only technology that illuminated our living spaces was the pear-shaped incandescent bulbs containing a thin wire inside. This technology, made practical by Thomas Edison, illuminated the world for more than 100 years. But there was a problem: These bulbs consumed a lot of electricity and became excessively hot compared to the light they provided.
To understand why an incandescent bulb is so inefficient, you don't need to flip through engineering books; the answer is hidden in the fundamental rule of physics that explains the emission of energy: Wien's Displacement Law.
How Does an Incandescent Bulb Work?
The working principle of incandescent bulbs is actually very simple: Inside a glass bulb that has been evacuated (or filled with an inert gas), there is a very thin wire (filament) with electrical resistance. This filament is usually made of tungsten metal, which has a very high melting point (around $3400^\circ\text{C}$).
When an electric current passes through this wire, the wire heats up to an extreme degree due to the resistance. When a metal is heated sufficiently, it begins to radiate just like glowing coals in a furnace (blackbody radiation). As the heat increases, some of the emitted photons enter the visible light spectrum, thus illuminating the room.
The Physics of the Problem: What Does Wien's Law Say?
Wien's Displacement Law states that the peak of the total thermal energy emitted by a heated body (the wavelength at which it is most intense) is inversely proportional to the absolute temperature of the body.
$$\lambda_{\text{max}} = \frac{b}{T}$$
In this formula, $b$ (Wien's constant) is approximately $2.898 \times 10^{-3} \text{ m}\cdot\text{K}$. $T$ is the temperature of the filament in Kelvin.
When a standard incandescent bulb filament is operating, its temperature is usually around $2700 \text{ K}$ to $3000 \text{ K}$ (about $2400^\circ\text{C}$ - $2700^\circ\text{C}$). Now let's calculate the peak wavelength for this temperature.
Taking the value $2800 \text{ K}$ as an example:
$$\lambda_{\text{max}} = \frac{2.898 \times 10^{-3}}{2800} \approx 1.035 \times 10^{-6} \text{ m} = 1035 \text{ nm (nanometers)}$$
Herein lies the fundamental problem at the heart of the bulb's inefficiency.
Visible Light vs. Infrared Heat
The human eye has evolutionarily adapted to the light of the Sun and can only "see" light between wavelengths of $400 \text{ nm}$ (violet) and $700 \text{ nm}$ (red).
The $1035 \text{ nm}$ value we calculated lies entirely outside the visible light spectrum, in the infrared (IR) region. In other words, the peak of the energy emitted by a standard incandescent bulb is not light, but heat. The bulb certainly emits radiation in the $400-700 \text{ nm}$ range as well (this is the yellowish light we see), but looking at the blackbody radiation curve (Planck curve), the vast majority of the energy (over 90%) is scattered around as invisible infrared waves.
Therefore, a 100-Watt incandescent bulb spends only about 5-10 Watts of its energy on lighting, and the remaining 90-95 Watts on heating the room (and your fingers if you try to touch it). For a device operated for lighting purposes to act like a heater 90% of the time is a massive inefficiency from an engineering standpoint.
Can't a Better Incandescent Bulb Be Made?
Looking at Wien's law, this idea immediately comes to mind: If we increase the temperature ($T$), the denominator of the formula will grow, so the peak wavelength ($\lambda_{\text{max}}$) will shrink and approach the visible region ($500-600 \text{ nm}$).
So why don't we raise the temperature to $5800 \text{ K}$ (the surface temperature of the Sun)? If we could do this, most of the energy would shift exactly into the visible white light region to which our eyes are most sensitive, and we would obtain a perfectly efficient lamp. However, here we hit the limits of materials science. No known solid material can remain in a solid state at a temperature of $5800 \text{ K}$. Even tungsten filaments melt around $3700 \text{ K}$. Although making the bulb burn a little brighter without melting (for example, raising it to the $3200 \text{ K}$ level as in halogen lamps) increases efficiency somewhat, it never offers a revolutionary solution.
This physical limit has led us to entirely new technologies that do not rely on thermal radiation (the blackbody principle) but directly convert electrical energy into photons, namely LEDs (Light Emitting Diodes).
Heat and Light Distribution in Different Lighting Technologies
We have detailed the thermal problems in Edison's incandescent bulb and how energy is lost as infrared radiation. So how is the situation in fluorescent and LED lamps?
Fluorescent lamps use an electric current to ionize a gas (usually mercury vapor). This ionization process produces ultraviolet (UV) light. However, the phosphor coating on the inside of the glass tube absorbs this invisible UV light and converts it into lower-energy visible white light. Since the blackbody radiation principle is not used in fluorescent lamps (i.e., a wire is not heated to very high temperatures), the inevitable infrared energy loss caused by Wien's law does not occur here.
LEDs (Light Emitting Diodes) use a completely different semiconductor technology. When an electric current is applied, electrons jump across energy gaps within the material, directly emitting photons at a specific wavelength (color). Because LEDs convert energy directly into the desired form of light without relying on heat production, they provide a massive efficiency advantage over incandescent bulbs. By directing all energy into the correct frequency band, we completely get rid of the "unwanted thermal infrared baggage" emitted by incandescent lamps.
Conduct Your Own Tests with the Wien's Displacement Law Calculator
To better understand the lighting characteristics of bulbs and the thermal properties of burning objects, you can use our Wien's Displacement Law Calculator tool.
Using the tool:
- Select the "Temperature to Wavelength" mode.
- Enter the value
3200in the temperature field for a halogen bulb and see for yourself where the peak is (approx. 905 nm). - Confirm why a charcoal fire (approx. $1200 \text{ K}$) only gives a very dull red light and emits its main energy in the 2.4-micrometer (infrared heat) band.
