At a temperature of 314 ~K and a pressure of 100 kPa , the speed of sound in a gas is 1380 ~ms ⁻¹ . The radius of each gas molecule is 0.5 Å . The frequency of sound at which the wavelength of sound wave in the gas becomes equal to the mean free path of the gas molecules is ( . Boltzmann constant =1.38 10⁻²³ JK ⁻¹ .)

Step-by-step solution

Given, temperature, T=314 ~K , pressure, p=100 kPa ,=1.0 10^5 ~Pa , speed of sound, v=1380 ~ms ⁻¹ and diameter of gas molecule, d=10⁻¹⁰ ~m or radius r= 1 2 10⁻¹⁰ ~m = 1 2 Å So, from Eqs. (i) and (ii), we get v= 2 d^2 p v k T Now, putting the given values, we get aligned & = 2 3.14 10⁻²⁰ 10^5 1380 1.38 10⁻²³ 314 v & = 2 10^9 ~Hz v=1000 2 MHz aligned Hence, the correct option is (b).