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.