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A black body radiates maximum power at a temperature of 5760 K. Then energy of photon emitted by the black body at maximum intensity is (Wien's constant is 0.288 K-cm)

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4.484 eV

2.480 eV

3.848 eV

2.884 eV

answer is A.

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Detailed Solution

$λ_{m} T = 0.288 \Rightarrow λ_{m} = \frac{0.288}{5760} cm = 0.00005 cm = 5000 \overset{o}{A}$

$∴$ Energy of photon $= \frac{hC}{λ} = \frac{12400}{5000} eV = 2.480 eV$

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