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Given that

Temperature T = 300 K

Atomic mass of nitrogen = 14.0076  amu

The molecular mass is the sum of the atomic masses of the two nitrogen atoms.

Nitrogen molecule mass m = 14.007 + 14.007 amu

                                         m = 28.014 * 1.66 * 10^{-27} kg                        \because \text{ 1 amu } = 1,66 * 10^{-27} kg

Step 1: Expression for speed of molecules (root mean square speed)  

              Required formulas:

                Kinetic energy K.E = \frac{1}{2} m(v_{rms})^2   ........................(1)

                           Where, m - mass and v_{rms} - root mean square speed

                Gaseous particle(nitrogen) kinetic energy at a certain temperature   K.E = \frac{3}{2} kT ................................(2)

                           Where,  K - Boltzmann constant 1.38*10^{-23} J/K and T - temperature

               From equation (1) and (2) we can write as follows

                           \frac{1}{2} m(v_{rms})^2 = \frac{3}{2} kT               

                                  (v_{rms})^2 = \frac{3kT}{m}

                                   v_{rms}=\sqrt{\frac{3kT}{m}} ..........................(3)             

Step 2: Equation for nitrogen de Broglie wavelength  

               de Broglie wavelength \lambda=\frac{h}{p}=\ \frac{h}{mv_{rms}}\ \ \ \ \ \ \ \ \ \ \

                     Where, h - Planck constant = 6.63* 10^{-34} J , p - linear momentum, m - mass and v_{rms} - speed  

                                                   \lambda = \frac{h}{m \sqrt{\frac{3kT}{m}}}                              \because \text{ Equation (3)}

                                                   \lambda = \frac{h}{ \sqrt{3kTm}}

Step 3:  Plug in the known value in the above equation  

                             \lambda = \frac{6.63*10^{-34}}{3 * 1.38*10^{-23} * 300 * 28.014 * 1.66 * 10^{-27} }

                             \lambda = 0.028 * 10^{-9} m

                             \lambda = 0.028 nm

                 Therefore, de Broglie wavelength of nitrogen molecule \lambda = 0.028 nm