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`]

Options

A. `10^(21)`
B. `10^(22)`
C. `10^(23)`
D. `10^(24)`

Answer

B

Explanation

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