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De Broglie wavelength (video) | Khan Academy


Bohr model energy levels (derivation using physics) ... Does the De Brogiie wavelength have any practical applications today? ... This means that a microscope using electron "matter waves" instead of photon light waves can see much smaller things. ... What am I supposed to imagine when I think about electrons now?


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Krishna
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a)

Mass of bullet m = 0.040 kg

Speed of a bullet v = 1.0 km/s= 10^3 m/s


          Momentum p = mv   

                    Where, p - momentum, m - mass and v - velocity  

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

                   Where, h - Planck constant = 6.63* 10^{-34} J   and p - linear momentum.    

          So, we can write de Broglie wavelength \lambda=\frac{h}{mv}


           Substitute the given values in the above equation              

                     \lambda = \frac{6.63* 10^{-34}}{0.040 * 10^3}

                     \lambda = 1.7 * 10^{-35} m


          Therefore, de Broglie wavelength    \lambda = 1.7 * 10^{-35} m



b)

Mass of ball m = 0.060

Speed of a ball v = 1.0 m/s

          de Broglie wavelength \lambda=\frac{h}{mv}

                     \lambda = \frac{6.63* 10^{-34}}{0.060* 1.0 }

                     \lambda = 1.1 * 10^{-32} m


c)

Mass of a dust particle m = 1.0 * 10^{-9} kg

Drifting speed of a particle v = 2.2 m/s

                de Broglie wavelength \lambda=\frac{h}{mv}

                     \lambda = \frac{6.63* 10^{-34}}{1.0 * 10^{-9} * 2.2}

                     \lambda = 3 * 10^{-25} m