I had posted this question once before, but no one helped me on this one. So I am posting this question again. Suppose the flame of a gas burner stove emits light just like a blackbody. The flame appears blue emiting most of the light at a wave length of 475 nanometers. What is the temperature of the flame? How do I solve this ?
To determine the temperature of the flame, you can use the concept of blackbody radiation and Wien's displacement law. This law states that the peak wavelength of radiation emitted by a blackbody is inversely proportional to its temperature.
Wien's displacement law can be expressed as:
λmax = b / T
Where:
- λmax is the peak wavelength of radiation (475 nm in this case)
- b is Wien's displacement constant (approximately equal to 2.9 × 10^-3 K·m)
- T is the temperature of the flame in Kelvin (what we're trying to find)
To solve for T, we can rearrange the equation:
T = b / λmax
Now, let's substitute the given values into the equation:
T = (2.9 × 10^-3 K·m) / (475 × 10^-9 m)
The units need to be compatible, so we multiply the numerator and denominator by 10^9 to convert meters to nanometers:
T = (2.9 × 10^-3 K·m) / (475 × 1)
T ≈ 6100 K
Therefore, the temperature of the flame is approximately 6100 Kelvin.
Remember, this result assumes that the flame behaves as a blackbody radiator, which might not be entirely accurate in practice.