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11 × 1011 Photons are incident on a surface in 10 s. These photons correspond to a wavelength of 10Å. If the surface area of the given surface is 0.01 m2, the intensity of given radiations is……. {h = 6.625 × 10–34 J. s, c = 3 × 108 (m/s)} 

(A) 21.86 × 10–3 (W / m2

(B) 2.186 × 10–3 (W / m2

(C) 218.6 × 10–3 (W / m2

(D) 2186 × 10–3 (W / m2)

1 Answer

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Best answer

The correct option is (B) 2.186 × 10–3 (W / m2).

Explanation:

Intensity of radiation = I = {(power) / (area)} = (P/A) = {E / (A ∙ t)} and E = {(nhc) / λ}

hence I = {(nhc) / (λAt)}

∴ I = [{11 × 1011 × 6.625 × 10–34 × 3 × 108} / {10 × 10–10 × 0.01 × 10}] ---- λ = 10 Å = 10 × 10–10m, A = 0.01 m2 and t = 10 sec

∴ I = 218.6 × 10–5

∴ I = 2.186 × 10–3 (W / m2) 

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