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A black body at 1227°emits radiation with maximum intensity at wave length of 5000 angstroms. The temperature of a body is increased by 1000°C . The maximum intensity will be observe at

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4000 angstroms

3000 angstroms

5000 angstroms

6000 angstroms

answer is D.

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Detailed Solution

$Given, T = 1227 + 273 = 1500 K$

$T^{'} = 1227 + 1000 + 273 = 2500 K$

$λ_{m} = 5000 Å$

$\frac{λ_{m}^{'}}{λ_{m}} = \frac{T}{T^{'}}$

$Hence, λ_{m}' = \frac{1500}{2500} \times 5000 = 3000 Å.$

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