Question

The number of quanta of radiation of frequency 4.98 ×10^14 s^-1 required to melt 100g of ice are

Answer 1

according to quantum mechanics, energy of each quanta , E = hv

where h is Plank's constant i.e., h = 6.63 × 10^-34 Js and v is frequency of radiation.

here given, frequency = 4.98 × 10¹⁴ s^-1

so, energy of each quanta , E = hv

= 6.63 × 10-³⁴ × 4.98 × 10¹⁴

≈ 33 × 10^-20 J

Let number of quanta = n required to melt 100g of ice.

now, latent heat of fusion of ice , L = 334 j/g

so, energy required to melt 100g = 334 × 100 J = 33400 J

now, energy lost by quantum radiation = energy gained by ice

or, n × 33 × 10^-20 = 33400

or, n = 33400/(33 × 10^-20) = 1012.12 × 10^20 = 1.01212 × 10²³ [ Ans]