Suppose a flame of a gas burner stove emits light just like a blackbody.The flame appears blue,emiting most of the light as a wave length of 475 nanometers. what is the temperature of the flame. How do I solve this?
You cannot get the flame temperature from that observation. You do not have blackbody continuum emission, which would be required to get a temperature from the spectrum. The wavelengths that you observe depends upon the emitting species, which happen to be trace amounts of C2 and CN.
T = 3 x 10^6 / wavelength
To solve this problem, we can use Wien's displacement law, which relates the wavelength at which a blackbody emits the maximum amount of radiation to its temperature.
The equation is:
λ_max = (b/T)
Where:
λ_max is the wavelength at which the blackbody emits maximum radiation
b is Wien's displacement constant (approximately 2898 μm.K)
T is the temperature of the blackbody in Kelvin
In your case, you are given that the flame emits most of the light at a wavelength of 475 nanometers (475 nm = 475 x 10^-9 m).
To find the temperature (T) of the flame, we need to convert the given wavelength into meters and then substitute it into the equation.
Here are the steps to solve it:
1. Convert the wavelength from nanometers to meters:
λ_max = 475 x 10^-9 m
2. Substitute the value of λ_max into Wien's displacement law equation:
475 x 10^-9 m = (b / T)
3. Rearrange the equation to solve for T:
T = b / (475 x 10^-9 m)
4. Substitute the value of Wien's displacement constant (b):
T = 2898 μm.K / (475 x 10^-9 m)
5. Convert the value of the displacement constant from micrometers to meters:
2898 μm = 2898 x 10^-6 m
6. Substitute the values into the equation and calculate T:
T = (2898 x 10^-6 m) / (475 x 10^-9 m)
Simplify the equation and convert the result to Kelvin (K) to find the temperature of the flame.