The energy need to liquify 10g of ice is 3.6kJ.If the radiation with frequency 4.53×10^14 Hz is used to liquify 5g of ice,the number of energy photons need for this is,? 1: 12×10^21 2: 6×10^21 3: 24×10^21 4: 18×10^24 5: 12×10^20

Well, first I'll verify the heat required...

#DeltaH_"fus" = "6.02 kJ/mol ice"#

#= "6.02 kJ"/cancel"mol ice" xx cancel"1 mol ice"/(18.015 cancel"g") xx 10.00 cancel"g" = "3.34 kJ"#

No matter if I round the numbers to the nearest whole number, this is what I get... So I'll proceed with this value... For #"5.00 g"# then we would involve #"1.17 kJ"# of thermal energy.

And so, if this was some laser that zapped ice,

#E_"laser" = "1170 J" = "Number of photons" xx E_"photon"#

For EACH photon we had according to what Planck proposed as a quantum of light energy:

#E_"photon" = hnu#

#= 6.626 xx 10^(-34) "J" cdot cancel"s"cdot"photon"^(-1) cdot 4.53 xx 10^14 cancel("s"^(-1))#

#= 3.002 xx 10^(-19) "J/photon"#

Energy, as you should know, is an extensive quantity. So, the ACTUAL number of photons in this red-hot laser is:

#color(blue)("Number of photons") = E_"laser"/E_"photon"#

#= (1170 cancel"J")/(3.002 xx 10^(-19) cancel"J""/photon") = ulcolor(blue)(3.90 xx 10^21 "photons")#

Verify that if we used your #DeltaH_"fus"#, we would instead get #6.00 xx 10^21 "photons"#.