$$\lambda_{max} = \frac{b}{T}$$
Where:
$\lambda_{max}$ is the wavelength of maximum emission intensity in meters
b is Wien's displacement constant (2.898 x 10^-3 m K)
T is the temperature of the blackbody in Kelvin
To determine the wavelength of maximum emission intensity for a blackbody with a temperature of 6000K, we simply plug the values into the equation:
$$\lambda_{max} = \frac{2.898 \times 10^{-3} \ m \ K}{6000 \ K} = 4.83 \times 10^{-7} \ m$$
Therefore, the wavelength of maximum emission intensity for a blackbody with a temperature of 6000K is 4.83 x 10^-7 m, which corresponds to the visible light spectrum (blue-green light).
