1.

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]

Answer»

`10^(21)`
`10^(22)`
`10^(23)`
`10^(24)`

Solution :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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