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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, ...

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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 ...

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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 ...

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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