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An electron, an alpha particle, and a proton have the same kinetic ...


An electron, an alpha particle, and a proton have the same kinetic energy. Which of these particles has the shortest de-Broglie wavelength? 7 Answers.


For more information, see An electron, an alpha particle, and a proton have the same kinetic ...

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I found an answer from en.wikipedia.org

Proton - Wikipedia


A proton is a subatomic particle, symbol p or p + , with a positive electric charge of +1e ... In previous years, Rutherford had discovered that the hydrogen nucleus (known ... particles, in the modern Standard Model of particle physics, protons are ... As a muon is 200 times heavier than an electron, its de Broglie wavelength is ...


For more information, see Proton - Wikipedia

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quantum mechanics | Definition, Development, & Equations ...


These properties include the interactions of the particles with one another and ... is converted into the kinetic energy meu2/2 of the emitted electron (me is the ... He found that part of the scattered radiation had the same wavelength λ0 as the ... radiation has both particle and wave characteristics, Louis-Victor de Broglie of  ...


For more information, see quantum mechanics | Definition, Development, & Equations ...

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I found an answer from opentextbc.ca

De Broglie's Matter Waves – University Physics Volume 3


Today, this idea is known as de Broglie's hypothesis of matter waves. In 1926, De Broglie's hypothesis, together with Bohr's early quantum theory, ... Any particle that has energy and momentum is a de Broglie wave of frequency f and wavelength \lambda : ... For matter waves, this group velocity is the velocity u of the particle.


For more information, see De Broglie's Matter Waves – University Physics Volume 3

Krishna
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Given that

Kinetic energy of electron =  Kinetic energy of alpha-particles = Kinetic energy of proton.


Step 1: Set up an equation for the kinetic energy in terms of 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}

    

Step 2: Get an expression for relation between the de Broglie wavelength and mass

             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{h}{\sqrt{2mK.E}}

                                                           \lambda \propto \frac{1}{\sqrt{m}}   


               For the same Kinetic-energy, the de Broglie wavelength associated with the particle is inversely proportional to the square

               root of its mass. Thus, the higher mass particle has the shortest wavelength.


               In this case, the protons are heavier than the electrons, and the alpha particles four times that of a proton. The alpha

               particle is also the heaviest compared to the proton and the electron. thus  alpha-particle has the shortest de Broglie

               wavelength.