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How many photos at 620 nm must be absorbed to melt `10 gm` of ice. If 320 J of heat is required to convert 1 gm of ice at `0^(@)C` [take : `hc = 1240 eV-nm`]A. `10^(21)`B. `10^(22)`C. `10^(23)`D. `10^(24)` |
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Answer» Correct Answer - B For melting 10 gm of ice Energy required `= 320 xx 10 = 3200 J` `E = n xx (1240)/(lambda)` E in eV, `lambda` in nm `(3200)/(1.6 xx 10^(-19)) = n xx (1240)/(620)` `n = (16000 xx 10^(-19))/(16) = 10^(22)` photons |
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