The temperature of our Sun’s surface is $~\mathbf{6000}\mathbf{K}$.(a) At what wavelength is the spectral emission of the Sun is maximum? (Refer to Exercise 79.) (b) Is there something conspicuous about this wavelength?

Given data can be listed below as:

The temperature of the sun is, $T=6000\text{K}$.$T=6000\text{K}$.

The law gives the relationship between the maximum wavelength of the radiation that was released from a black body and the temperature at which the black body was kept; this relationship can be expressed as,

${\mathit{\lambda }}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathit{T}\mathbf{=}\mathit{b}$ (1)

Here ${\mathit{\lambda }}_{\mathbf{m}\mathbf{a}\mathbf{x}}$is the maximum wavelength, localid="1660183413143" $\mathit{T}$is the temperature at which the black body was kept, and localid="1660183425059" $\mathit{b}$is known as Wien’s displacement constant.

Let’s modify equation 1 as,

$\begin{array}{rcl}{\lambda }_{\max }& =& \frac{b}{T}\\ & & \end{array}$ (2)

The value of Wien’s displacement constant is, $b=2.897\times {10}^{-3}\text{m}·\text{K}$.

The wavelength at which the spectral emission of the Sun is maximum can be calculated from the above expression. The maximum spectral emission is associated with the maximum wavelength (peak in the energy-temperature curve) for that given temperature.

Substitute the values in equation 2, and we get,

$\begin{array}{rcl}{\lambda }_{\max }& =& \frac{2.897\times {10}^{-3}\text{m}·\text{K}}{6000\text{K}}\\ & =& 4.828\times {10}^{-7}\text{m}\\ & & \end{array}$

Thus, the wavelength at which the spectral emission is maximum is $4.828\times {10}^{-7}\text{m}$.

From the calculation of part a, the wavelength at which the spectral emission is maximum is $4.828\times {10}^{-7}\text{m}$.

The visible spectrum exists between $300\text{nm}$and $700\text{nm}$.

The wavelength we calculated in nm can be calculated as,

$\begin{array}{rcl}& =& 4.828\times {10}^{-7}\text{m}\\ & =& 4.828\times {10}^{-7}·\left(1\text{m}\times \frac{{10}^9\text{nm}}{1\text{m}}\right)\\ & =& 4.828\times {10}^{-7}\times {10}^9\text{nm}\\ & =& 4.828\times {10}^2\text{nm}\\ & =& 482.8\text{nm}\end{array}$

This is between the visible range in the middle, to be precise.

Thus, it is conspicuous as it is in the visible range of the spectrum.