Wien’s law is stated as follows: λ_mT = C, where C is 2898 μm.K and λ_m is the wavelength at which the emissive power of a black body is maximum for a given temperature T. The spectral hemispherical emissivity (ε_λ) of a surface is shown in the figure below (1 Å = 10^-10 m). The temperature at which the total hemispherical emissivity will be highest is K (round off to the nearest integer).

The correct answer is 4830.

Step-by-Step Explanation:

Wien's displacement law relates the temperature $T$ of a black body to the wavelength ${\lambda }_m$ at which its spectral emissive power is maximum. The law is expressed as:
${\lambda }_{m}T=C$
where the constant $C=2898\text{μm·K}$.

From the spectral hemispherical emissivity (${\epsilon }_{\lambda }$) distribution shown in the figure, the emissivity reaches its highest value at a specific wavelength of:
$\lambda =6000\text{Å}$
To find the temperature at which the total hemispherical emissivity is highest, we align the peak wavelength of black body radiation ${\lambda }_m$ with this wavelength of maximum spectral emissivity.

First, we convert the wavelength from Angstroms (Å) to micrometers (μm) to match the units of the Wien's constant $C$:
$1\text{��}={10}^{-10}\text{m}$
${\lambda }_m=6000\times {10}^{-10}\text{m}$
Since $1\text{μm}={10}^{-6}\text{m}$:
${\lambda }_m=\frac{6000\times {10}^{-10}\text{m}}{{10}^{-6}\text{m/μm}}=0.6\text{μm}$

Using Wien's displacement law, we can solve for the corresponding temperature $T$:
${\lambda }_{m}T=C$
$0.6\text{μm}\times T=2898\text{μm·K}$
$T=\frac{2898}{0.6}=4830\text{K}$

Thus, the temperature at which the total hemispherical emissivity will be highest is 4830 K.