A light of wavelength 200 nm falls upon a surface and two different wave length photons $λ = 800 n m$ and $λ = 400 n m$ are emitted from the surface. 80% of the energy absorbed is re-emitted in the form of photon. Number of photons emitted as $λ = 800 n m$ is three times than that of number of photons emitted as $λ = 400 n m .$ If the ratio of the total absorbed photon to total emitted photon is x, then find out the numerical value of 64x?

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answer is 25.

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Detailed Solution

: Let no. of photons with wave length 200 $n m = n_{1}$
No of photos with wavelength 800 $n m = n_{2}$
No of photons with wavelength 400 $n m = n_{3}$
$x = \frac{n_{1}}{n_{2} + n_{3}} ..... (1) [n_{2} \frac{h c}{λ_{2}} + n_{3} . \frac{h c}{λ_{3}}] = 0.8 n_{1} . \frac{h c}{λ_{1}} \frac{n_{2}}{λ_{2}} + \frac{n_{3}}{λ_{3}} = \frac{0.8 n_{1}}{λ_{1}} ..... (2) A l s o n_{2} = 3 n_{3} ...... (3) x = \frac{5}{12.8} ∴ 64 x = 25$