The energy required to melt 1g. ice is 33J. The number of quanta of radiation of frequency 4.67 × 10¹³ sec^–1 that must be absorbed to melt 10g ice is

$$\begin{gathered} \mathrm{Energy} \mathrm{required} \mathrm{to} \mathrm{melt} \mathrm{one} \mathrm{gram} \mathrm{of} \mathrm{ice} = 33\mathrm{J} \\ \mathrm{Energy} \mathrm{required} \mathrm{to} \mathrm{melt} \mathrm{ten} \mathrm{grams} \mathrm{of} \mathrm{ice} = 330\mathrm{J} \\ \mathrm{Energy} \mathrm{of} \mathrm{each} \mathrm{quanta} = \mathrm{hv}= (6.625 \times {10}^{-34}\mathrm{J}.\mathrm{S})(4.67 \times {10}^{13}{\mathrm{s}}^{-1}) \\ = (30.93 \times {10}^{-21}\mathrm{J}) \\ \mathrm{Total} \mathrm{energy} = \mathrm{Nhv} (\mathrm{N} = \mathrm{no}. \mathrm{of} \mathrm{quanta}) \end{gathered}$$

No.of Quanta N=$\frac{\mathrm{Total} \mathrm{energy} \mathrm{required} \mathrm{to} \mathrm{melt} \mathrm{in} \mathrm{gms} \mathrm{of} \mathrm{ice}}{\mathrm{Energy} \mathrm{of} \mathrm{each} \mathrm{quanta}}$

 N= $\frac{ 330}{30.93 \times {10}^{-21}}$ =1.065 × 10²²