Question

Calculate the de Broglie wavelength for (a) an electron with a kinetic energy of 100eV, (b) a proton with a kinetic energy of 100 eV, and (c) an electron in the first Bohr orbit of a hydrogen atom.

Answer 1

Answer:

Broglie wavelength: electron 1.22 10⁻¹⁰ m , proton 2.87 10⁻¹² m , hydrogen atom 7.74 10⁻¹² m

Explanation:

The equation given by Broglie relates the momentum of a particle with its wavelength.

       p = h /λ

In addition, kinetic energy is related to the amount of movement

      E = ½ m v²

      p = mv

      E = ½ p² / m  

      p = √2mE

If we clear the first equation and replace we have left

       λ = h / p =

       λ = h / √2mE

Let's reduce the values ​​that give us SI units

      1 ev = 1,602 10⁻¹⁹  J

      E1 = 100 eV (1.6 10⁻¹⁹ J / 1eV) = 1.6 10⁻¹⁷ J

We look in tables for the mass of the particle and the Planck constant

      h = 6,626 10-34 Js

      me = 9.1 10-31 Kg

      mp = 1.67 10-27 Kg

Now let's replace and calculate the wavelengths

a) Electron

       λ1 = 6.6 10⁻³⁴ / √(2 9.1 10⁻³¹ 1.6 10⁻¹⁷) = 6.6 10⁻³⁴ / 5.39 10⁻²⁴

       λ1 = 1.22 10⁻¹⁰ m

b) Proton

       λ2 = 6.6 10-34 / √(2 1.67 10⁻²⁷ 1.6 10⁻¹⁷) = 6.6 10⁻³⁴ / 2.3 10⁻²²

       λ2 = 2.87 10⁻¹² m

c) Bohr's first orbit

       En = 13.606 / n2 [eV]

       n = 1

       E1 = 13.606 eV

       E1 = 13,606 ev (1.6 10⁻¹⁹ / 1eV) = 21.77 10⁻¹⁹ J

       λ3 = 6.6 10⁻³⁴ /√(2 1.67 10⁻²⁷ 21.77 10⁻¹⁹) = 6.6 10⁻³⁴ / 8.52 10⁻²³

       λ3 = 0.774 10⁻¹¹ m = 7.74 10⁻¹² m