Answer:
The de Broglie wavelength of electron βe = 2.443422 × 10⁻⁹ m
The de Broglie wavelength of proton βp = 5.70 × 10⁻¹¹ m
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
