For an electron with KE = 1 eV and rest mass energy 0.511 MeV, the associated DeBroglie wavelength is 1.23 nm, about a thousand times smaller than a 1 eV photon. Relation Between de Broglie Equation and Bohr’s Hypothesis of Atom. It has practically zero wave characteristics. The kinetic energy is given as 1 eV. Answer: The de Broglie wavelength of the photon can be found using the formula: λ = 4.42 x 10 (-7) m. λ = 442 x 10 (-9) m. λ = 442 nm. This equation relating the momentum of a particle with its wavelength is de Broglie equation and the wavelength calculated using this relation is de Broglie wavelength. 2) The de Broglie wavelength of a certain electron is . Energy of proton, neutron and electron = constant. The De Broglie wavelength equation is as follows: λ=h/p. (This is why the limiting resolution of an electron microscope is much higher than that of an optical microscope.) The kinetic energy being the same for all, thus the only thing left to compare is the √mass. m is the object's mass in "kg" and v is its velocity in "m/s". De-Broglie wavelength is calculated by using the formula: where, = wavelength of electron h = Planck's constant = m = mass of electron = v = velocity of electron = 1% of c= The de broglie wavelength of an electron moving with 1% of the speed of light 2.41 Angstrom. De- Broglie wave length is given by - λ=h/√2mk. Kinetic energy of particle of mass m having momentum p is, K.E = 12p2m ⇒ p = 2mK De-Broglie wavelength, λ = hp = h2mK ∴ p = hλ ....(i)and, K = h22mλ2 ....(ii) If λ is constant, then from equation (i), p = constant. Bohr postulated that angular momentum of an electron revolving around the nucleus as quantized. The de-Broglie Wavelength is inversely proportional to √ of mass and the kinetic energy. When k is constant = λα/√1m. Remember 1 eV is equal to 1.6 x 10-19 Joules. As per De-broglies formula, Kinetic energy of proton is equal to kinetic energy of proton. The de Broglie wavelength of the photon is 442 nm. K=1/2mv² mv=√2ml. We know the value of Planck's constant h and so to calculate the wavelength all we need is the momentum, which is equal to mv. Since, mass of proton > mass of electron, This implies, That is, wavelength of electron is greater than the wavelength of proton. The de Broglie wavelength lambda in units of "m" for mass-ive objects is given by the de Broglie relation: lambda = h/(mv), where: h = 6.626 xx 10^(-34) "J"cdot"s" is Planck's constant. This wavelength is in the blue-violet part of the visible light spectrum.

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