Question

What is the average de Broglie wavelength of a gas of nitrogen molecules at a temperature of 300K? Assume that the gas is ideal and is at 1 atm of pressure. Calculate the interatomic spacing. What would the temperature have to be for the same amount of nitrogen gas in the same volume to have a de Broglie wavelength equal to the interatomic spacing?

Answer 1

Is Average de Broglie wavelength of the gas of nitrogen molecule, We know Atomic mass of nitrogen = 14.00760 mass of No molecule = 2x14.0076 - 27 +2 x 14.0076 X 166 x 10 kg -23 5xlo [: 1U = 1.66 X10 kg] mass (m) = 4.61 Plank's constant (h) = 6.63 x 1034 J sec Boltzmann Constant (k)= 138x133Jk'. kinetic energy of the molecule, KE = 1 xm x Wrms? kinetic energy of gas molecule, K. E= by equating both the energies, m 189ms)' = Z KT vorms - 3*T Now, de-b109 lie wavelength, Am Vorms ml BKT 3KT h 3k Т U m h - To mk

6:63 x 1034 J 3x4.65X1026 x 1.38×10 23 x 300 Here To 300k (given) - 13.11 ) x 1025 = 2.75X10 m. 2), Now, At 1 atm psiesswe, you interatomie spacing, We have the ideal gas formula, PV =RT PV = KNT [R=KN] Where v = volume of the gas. N = Number of moles of the gas a Interatomice spacing of the gas, 4= ☺ - (T) r=[ 1.38X1023x3007 18105 M = 3.35 x10 m.) We have to find the d= 3.35x10 m. temperature 3- Now at

We know, 3 m KT On squarung on both sides, +²= h² 3 m kТ n² 3 mK T = ya d² x 3 mK T= (6.63 x10 34) 2 (3.35x10 % 3x4.6581026 x 1.38x123 6.63 X 6.63 x 10 3.35 X3.35 x 4.65X3 X138 x1067 T= 43.956 x 10 216.044 T= 0.203x10 T= 0·0203K /