Can anuone help with part 2? PART 1: How many photons at 660 nm must be absorbed
ID: 688201 • Letter: C
Question
Can anuone help with part 2? PART 1: How many photons at 660 nm must be absorbed to melt4.0 x102 g ofice? (Hint: It takes 334 J to melt 1 g of ice at 0°C.) ANSWER:4.44e23 photons. PART 2: On average, howmany H2O molecules does one photon convert from ice towater? Can anuone help with part 2? PART 1: How many photons at 660 nm must be absorbed to melt4.0 x102 g ofice? (Hint: It takes 334 J to melt 1 g of ice at 0°C.) ANSWER:4.44e23 photons. PART 2: On average, howmany H2O molecules does one photon convert from ice towater?Explanation / Answer
Given that - 1 g of ice take 334 J of energy to melt in to water =4.0*102 Thus the amount of energy required to melt the 400 g ofice = 400 g * 334 J /1 g = 133600 J The wave length corresponds to this enery = E = hc / 133600 J =6.0*10-34 J.sec * 3.0*108 m/sec / Thus wave length, = 1.34*10-30m Hence the number of photons that are absorbed at 660 nm= 1.34*10-30 m / 660*10-9 m =2.03*10-24 nm b) 2.03*10-24 of photonsconverts the 400 g*6.023*1023 /18.0 g/mol ofH2O molecules, hence - 1 photon converts the number of H2Omolecules = 400 g*6.023*1023 /18.0g/mol *2.03*10-24 = 6.593 Thus the amount of energy required to melt the 400 g ofice = 400 g * 334 J /1 g = 133600 J The wave length corresponds to this enery = E = hc / 133600 J =6.0*10-34 J.sec * 3.0*108 m/sec / Thus wave length, = 1.34*10-30m Hence the number of photons that are absorbed at 660 nm= 1.34*10-30 m / 660*10-9 m =2.03*10-24 nm b) 2.03*10-24 of photonsconverts the 400 g*6.023*1023 /18.0 g/mol ofH2O molecules, hence - 1 photon converts the number of H2Omolecules = 400 g*6.023*1023 /18.0g/mol *2.03*10-24 = 6.593Related Questions
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