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Calculate the de-Broglie wavelength of an electron acelerated from ...


According to Louis de Broglie's famous equation, Wave Nature (Wavelength) ... Momentum = sqrt(2 x Mass x Kinetic Energy) [ As K.E. = 0.5mv^2, and ... The question isn't quite right, potential difference is specified in V (volt) not eV ( electron volt). ... If an electron is accelerated from rest through a potential difference of 10V, ...


For more information, see Calculate the de-Broglie wavelength of an electron acelerated from ...

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Electron - Wikipedia


De Broglie's prediction of a wave nature for electrons led Erwin Schrödinger to postulate a wave equation for electrons moving under the influence of the nucleus ...


For more information, see Electron - Wikipedia

Qalaxia Master Bot
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De Broglie wavelength (video) | Khan Academy


Bohr model energy levels (derivation using physics) ... Since electrons have a rest mass, unlike photons, they have a de Broglie wavelength ... This means that a microscope using electron "matter waves" instead of photon light waves can see ... For example in the double slit experiment.... electron waves diffract and then ...


For more information, see De Broglie wavelength (video) | Khan Academy

Krishna
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Given that

Potential difference(voltage) V = 56 V


a) Calculate the momentum

            Recall the formula of momentum and kinetic energy

                  Momentum p = mv

                  Kinetic energy K.E = \frac{1}{2} mv^2

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


                     K.E = \frac{1}{2} m * (\frac{p}{m})^2                 \because p=mv


                     K.E = \frac{p^2}{2m}


                     p = \sqrt{2mK.E}

    

                  According to the conversion law

                          K.E = eV

                  Where, charge of electron e = 1.6* 10^{-19} and V - potential difference(voltage)    


                  Therefore, momentum p = \sqrt{2meV}


                  Substituting the known values in the above equation

                              Mass of an electron = 9.1 * 10^{-31}   kgs

                               p = \sqrt{2 * 9.1 * 10^{-31} * 1.6*10^{-19} V * 56}

                               p = 4.03 *10^{-24} kg m/s

          

                Hence, Momentum p = 4.03 *10^{-24} kg m/s


b) de Broglie wavelength of an electron

        Recall the de Broglie wavelength formula

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

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

                             \lambda = \frac{6.63*10^{-34}}{4.03*10^{-24}}       

                             \lambda = 1.64*10^{-10} m

                             \lambda = 0.164 nm


                  Therefore, de Broglie wavelength of an electron   \lambda = 0.164 nm