Question

the energy required to melt 1g of ice is 33j the no.of quanta of radiation of frequency 4.67x10-13 sec that must be absorbed to melt 10g ice​

Answer 1

Answer:

1 g requires 33 J

so 10 g will require 330 J heat

Total quanta=E=hv=$6.62*10^{-34} *4.67*10^{14}$

                              =$30.91*10^{-20}$

10 g of ice will require=$\frac{330}{30.91*10^{-20} }$

                                   =$1.065*10^{22}$

Answer 2

Answer:1.067*10^22

Explanation:Energy required to melt 1g ice =33J.

Energy required to melt 10g ice =33×10=330J.

Radiation frequency (ν)=4.67×10

13

s

−1

As we know, the energy of radiation is given by:

E=hν

Here, h= Planck's constant =6.62×10

−34

J.s

∴E=6.62×10

−34

×4.67×10

13

⇒E=3.09×10

−20

J

∴ No. of quanta required =

3.09×10

−20

330

​

=1.067×10

22