Final answer:
To find the temperature of gas near a black hole emitting UV radiation at 3.6 nm, we use Wien's Displacement Law. The temperature is inversely related to the peak wavelength, and by substituting the given values into Wien's formula, the estimated temperature of the gas is roughly 800,000 K.
Step-by-step explanation:
The question given involves finding the temperature of a gas emitting ultraviolet radiation at a wavelength characteristic of a blackbody spectrum.
To determine the temperature of the gas near a black hole emitting radiation at a wavelength of 3.6 nm, one would use Wien's Displacement Law which states that the blackbody radiation curve for different temperatures will peak at wavelengths inversely proportional to the temperature.
The formula is λ_max = b/T, where λ_max is the peak wavelength, T is the temperature in Kelvin, and b is Wien's displacement constant approximately equal to 2.897 x 10^-3 m·K. By rearranging the formula to solve for T, we find T = b / λ_max. Substituting the given wavelength of 3.6 nm (3.6 x 10^-9 meters) into the equation, we calculate the temperature of the gas.
Therefore, the temperature (T) of the gas would be approximately T = (2.897 x 10^-3 m·K) / (3.6 x 10^-9 m) = 8.04 x 10^5 K, or roughly 800,000 K.