Question

Calculate the value of λmax for solar radiation assuming that surface temperature of Sun is 5800 K (b=2.897×10^-3 mK).​

Answer 1

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★GIVEN:-

• Temperature of Sun = 5800K

• λmaxT = constant b (2.897 × 10^-3mK)

★TO FIND:-

• λmax

★FORMULA USED:-

• $\displaystyle\sf \bf{T = \frac{b}{λmax}}$

★EXPLANATION:-

According to Wein's law,

λmaxT = Constant b = 2.892 × 10^-3mK

Hence,

$\displaystyle \sf \bf{ \frac{b}{T} = λmax}$

$\displaystyle \sf \bf{λmax = \frac{2.897× 10^-3mK}{5800K } }$

$\displaystyle \sf \bf{λmax = 4.99 × 10^-7m}$

In Å

$\displaystyle \: { \boxed{ \underline{ \color{teal}{ \sf{ \bf{λmax = 4990Å}}}}}}$

So, the λmax for solar radiation will be 4990Å (approx).

Answer 2

Given :

T = 5800 K

$\sf b = 2.897 × 10^{-3}$ mK

To find :

$\sf \lambda_{max}$ = ?

Solution :

Wien's displacement law : The wavelength $\sf \lambda_{max}$ corresponding to maximum energy emission by a black body at absolute temperature T is given by $\sf \lambda_{max} = \dfrac{b}{T}$.

• $\sf \lambda_{max} = \dfrac{2.897 × 10^{-3} m K}{5800 K}$

• $\sf \lambda_{max} = 0.0004994 × 10^{-3} m$

• $\sf \lambda_{max} = 4.994 × 10^{-3}m$

• $\sf \lambda_{max}$ = 4994 Å

The value of $\sf \lambda_{max}$ for solar radiation is 4994 Å.