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.