Answer
Applying Wein's displacement
[tex]\lamda_{max}\ T = 2898 \mu_mK[/tex]
1) for sun T = 5800 K
[tex]\lambda_{max} = \dfrac{2898}{5800}[/tex]
[tex]\lambda_{max} = 0.5 \mu_m[/tex]
2) for tungsten T = 2500 K
[tex]\lambda_{max} = \dfrac{2898}{2500}[/tex]
[tex]\lambda_{max} = 1.16 \mu_m[/tex]
3) for heated metal T = 1500 K
[tex]\lambda_{max} = \dfrac{2898}{1500}[/tex]
[tex]\lambda_{max} = 1.93 \mu_m[/tex]
4) for human skin T = 305 K
[tex]\lambda_{max} = \dfrac{2898}{305}[/tex]
[tex]\lambda_{max} = 9.50 \mu_m[/tex]
5) for cryogenically cooled metal T = 60 K
[tex]\lambda_{max} = \dfrac{2898}{60}[/tex]
[tex]\lambda_{max} = 48.3 \mu_m[/tex]
range of different spectrum
UV ----0.01-0.4
visible----0.4-0.7
infrared------0.7-100
for sun T = 5800
λ 0.01 0.4 0.7 100
λT 58 2320 4060 5.8 x 10⁵
F 0 0.125 0.491 1
fractions
for UV = 0.125
for visible = 0.441-0.125 = 0.366
for infrared = 1 -0.491 = 0.509