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Question: The temperature of an ideal gas in 3- deimensions is 300 K. The corresponding de - Broglie wavelengt...

The temperature of an ideal gas in 3- deimensions is 300 K. The corresponding de - Broglie wavelength of the electron approximately at 300 K , is : [ m e = mass of electron = 9 × 10 − 31 k g h = Planck constant = 6.6 × 10 − 34 J s k B = Boltzmann constant = 1.38 × 10 − 33 J K − 1 ]

A

6.24 × 10-9 m

B

6.24 × 10-4 m

C

1.057 × 10-25 m

D

6.6 × 10-34 m

Answer

6.24 × 10-9 m

Explanation

Solution

The de Broglie wavelength (λ\lambda) of a particle is given by λ=hp\lambda = \frac{h}{p}. For a particle with mass mm and kinetic energy KEKE, the momentum is p=2m⋅KEp = \sqrt{2m \cdot KE}. Thus, λ=h2m⋅KE\lambda = \frac{h}{\sqrt{2m \cdot KE}}. The average translational kinetic energy of a particle in a 3D ideal gas at temperature TT is KEavg=32kBTKE_{avg} = \frac{3}{2} k_B T. Substituting this into the de Broglie wavelength formula, we get λ=h2m⋅32kBT=h3mkBT\lambda = \frac{h}{\sqrt{2m \cdot \frac{3}{2} k_B T}} = \frac{h}{\sqrt{3m k_B T}}.

Given values:

  • Mass of electron, me=9×10−31m_e = 9 \times 10^{-31} kg
  • Planck constant, h=6.6×10−34h = 6.6 \times 10^{-34} J s
  • Temperature, T=300T = 300 K
  • Boltzmann constant, kB=1.38×10−23k_B = 1.38 \times 10^{-23} J K−1^{-1} (assuming the provided 10−3310^{-33} was a typo and the standard value is used).

Calculate 3mekBT3m_e k_B T: 3×(9×10−31 kg)×(1.38×10−23 J K−1)×(300 K)=1.1178×10−50 kg J3 \times (9 \times 10^{-31} \text{ kg}) \times (1.38 \times 10^{-23} \text{ J K}^{-1}) \times (300 \text{ K}) = 1.1178 \times 10^{-50} \text{ kg J}

Calculate the momentum term: 1.1178×10−50 kg J≈1.057×10−25 kg m/s\sqrt{1.1178 \times 10^{-50} \text{ kg J}} \approx 1.057 \times 10^{-25} \text{ kg m/s}

Calculate the de Broglie wavelength: λ=6.6×10−34 J s1.057×10−25 kg m/s≈6.24×10−9 m\lambda = \frac{6.6 \times 10^{-34} \text{ J s}}{1.057 \times 10^{-25} \text{ kg m/s}} \approx 6.24 \times 10^{-9} \text{ m}