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Thermal Properties Of Matter

Question

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A black body has a wavelength of A at temperature 2000 K. Its corresponding wavelength at temperature 3000 K will be

$\frac{2 λ}{3}$

$\frac{3 λ}{2}$

$\frac{4 λ}{9}$

$\frac{9 λ}{4}$

Solution

$\frac{2 λ}{3}$

Below figure gives the experimental curves for radiation energy per unit area per unit wavelength emitted by blackbody versus wavelength for different temperatures.

The wavelength λ_m for which energy is maximum decreases with increasing temperature. The relation between λ_m and T is given by Wein's Displacement Law

According to Wein's displacement law,

λ_m T = Constant

⇒ $\frac{λ_{2}}{λ_{1}} = \frac{2000}{3000}$

$\frac{λ_{2}}{λ} = \frac{2}{3}$

$λ_{2} = \frac{2}{3} λ$

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Thermal Properties Of Matter

A black body has a wavelength of A at temperature 2000 K. Its corresponding wavelength at temperature 3000 K will be

$\frac{2 λ}{3}$

$\frac{3 λ}{2}$

$\frac{4 λ}{9}$

$\frac{9 λ}{4}$

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For Daily Free Study Material Join wiredfaculty

Thermal Properties Of Matter

A black body has a wavelength of A at temperature 2000 K. Its corresponding wavelength at temperature 3000 K will be 2λ3 3λ2 4λ9 9λ4

Some More Questions From Thermal Properties of Matter Chapter

A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

A black body has a wavelength of A at temperature 2000 K. Its corresponding wavelength at temperature 3000 K will be

$\frac{2 λ}{3}$

$\frac{3 λ}{2}$

$\frac{4 λ}{9}$

$\frac{9 λ}{4}$