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Bohr model energy levels (derivation using physics) ... Does the De Brogiie wavelength have any practical applications today? ... For example in the double slit experiment.... electron waves diffract and then 'condense' onto the ... the photoelectric effect by saying that the light is a wave that carries energy, and this energy ...


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Veda
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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