Question

An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a temperature of 1950 K. (kb is Boltzmann's constant, 1.38x10-23 J/K).

Answer 1

The de Broglie wavelength of electron βe = 2.443422 × 10⁻⁹ m

The de Broglie wavelength of proton βp = 5.70 × 10⁻¹¹ m

Step-by-step explanation:

Thermal kinetic energy of electron or proton = KE

∴ KE = 3kbT/2

given that; kb = 1.38 x 10⁻²³ J/K , T = 1950 K

so we substitute

KE = ( 3 × 1.38 x 10⁻²³ × 1950 ) / 2

kE = 4.0365 × 10⁻²⁰ ( is the kinetic energy for both electron and proton at temperature T )

Now we know that

mass of electron M'e = 9.109 × 10⁻³¹

mass of proton M'p = 1.6726 × 10⁻²⁷

We also know that

KE = p₂ / 2m

from the equation, p = √ (2mKE)

{ p is momentum, m is mass }

de Broglie wavelength = β

so β = h / p = h / √ (2mKE)

h = Planck's constant = 6.626 × 10⁻³⁴

∴ βe = h / √ (2m'e × KE)

βe = 6.626 × 10⁻³⁴ / √ (2 × 9.109 × 10⁻³¹ × 4.0365 × 10⁻²⁰ )

βe = 6.626 × 10⁻³⁴ / √ 7.3536957 × 10⁻⁵⁰

βe = 6.626 × 10⁻³⁴ / 2.71176984642871 × 10⁻²⁵

βe = 2.443422 × 10⁻⁹ m

βp = h / √ (2m'p ×KE)

βp = 6.626 × 10⁻³⁴ / √ (2 × 1.6726 × 10⁻²⁷ × 4.0365 × 10⁻²⁰ )

βp = 6.626 × 10⁻³⁴ / √ 1.35028998 × 10⁻⁴⁶

βp = 6.626 × 10⁻³⁴ / 1.16201978468527 × 10⁻²³

βp = 5.702140 × 10⁻¹¹ m