Question

Under daylight conditions, the human eye exhibits maximum sensitivity to light at a wavelength of \(555 \mathrm{~nm}\). At what temperature (K) would a blackbody emit light with a peak intensity at this wavelength?

Short answer

Answer: The temperature of a blackbody that emits light with a peak intensity at a wavelength of 555 nm is approximately 5210 K.

Step-by-step solution

Convert the given wavelength to meters

The wavelength is given in nanometers, we need to convert it to meters to use in Wien's Law: $$\lambda_{max}=555\mathrm{~nm}\times\frac{1~\mathrm{m}}{10^9~\mathrm{nm}}=5.55\times10^{-7}\mathrm{~m}$$

Rearrange Wien's Law formula to isolate the temperature (T)

To find the temperature, we need to rearrange the Wien's Law formula to get T on one side: $$T=\frac{b}{\lambda_{max}}$$

Substitute the values into the formula

Now we can substitute the values of \(\lambda_{max}\) and b into the formula to find the temperature: $$T=\frac{2.898\times10^{-3}~\mathrm{m\cdot K}}{5.55\times10^{-7}~\mathrm{m}}$$

Calculate the temperature

Divide the values to find the temperature of the blackbody: $$T\approx\frac{2.898\times10^{-3}}{5.55\times10^{-7}}\approx5214.41~\mathrm{K}$$

Round off the answer

Round off the answer to a reasonable number of significant digits. In this case, round it to 3 significant digits: $$T\approx5210~\mathrm{K}$$ The temperature of a blackbody that emits light with a peak intensity at a wavelength of \(555 \mathrm{~nm}\) is approximately \(5210~\mathrm{K}\).