(A) Given: Surface temperature $T = 6000 \,K$,Wien's constant $b = 2.897 \times 10^{-3} \,m-K$.According to Wien's displacement law,$\lambda_{max} T =

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(A) Given: Surface temperature $T = 6000 \,K$,Wien's constant $b = 2.897 \times 10^{-3} \,m-K$.According to Wien's displacement law,$\lambda_{max} T = b$.Therefore,$\lambda_{max} = \frac{b}{T} = \frac{2.897 \times 10^{-3}}{6000} \,m$.$\lambda_{max} = 4.828 \times 10^{-7} \,m$.Since $1 \,Å = 10^{-10} \,m$,we have $4.828 \times 10^{-7} \,m = 4828 \times 10^{-10} \,m = 4828 \,Å$.

Class 11 Physics 10-2.Heat Transfer Spectral Emissive Power and Wien's Displacement Law

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Assume that for solar radiation,the surface temperature of the Sun is $6000 \,K$. If Wien's constant '$b$' is $2.897 \times 10^{-3} \,m-K$,the value of the maximum wavelength will be: (in $Å$)

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A
$4828$
B
$3648$
C
$6400$
D
$0.18$

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10-2.Heat Transfer

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Spectral Emissive Power and Wien's Displacement Law

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Assume that for solar radiation,the surface temperature of the Sun is $6000 \,K$. If Wien's constant '$b$' is $2.897 \times 10^{-3} \,m-K$,the value of the maximum wavelength will be: (in $Å$)

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Wien's displacement law expresses the relationship between:

Two stars $A$ and $B$ radiate maximum energy at wavelengths $3.9 \times 10^{-7} \text{ m}$ and $5.2 \times 10^{-7} \text{ m}$ respectively. The ratio of the temperature of star $A$ to that of star $B$ will be

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