Question

enclosure fill with helium is heated at 400kelvin and beam of helium atom out of enclosure. calculate the De-broglie wavelength . corresponding to helium atom given that mass of helium is 6.7 * 10^-27​

Answer 1

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Answer 2

Given info : An enclosure filled with helium is heated to a temperature of 400 K. A beam of helium emerges out of the enclosure. mass of helium atom is 6.7 × 10^-27 kg

To find : mean De-Broglie's wavelength is ..

solution : using formula,

first find mean velocity of helium atom, v_m = √{8kT/πm}

here, Boltzmann's constant, k = 1.38 × 10^-23 J/K

Temperature, T = 400K

mass of helium atom, m = 6.7 × 10^-27 kg

= √{8 × 1.38 × 10^-23 J/K × 400K/3.14 × 6.7 × 10^-27}

= √(8 × 1.38 × 4 × 10⁶/3.14 × 6.7)

= 1.4488 × 10³ m/s

= 1448.8 m/s

now using formula of De-Broglie's wavelength

λ = h/mv_m

here, h = 6.63 × 10^-34 Js

m = 6.7 × 10^-27 kg

v_m = 1448.8 m/s

so, λ = (6.63 × 10^-34)/(6.7 × 10^-27 × 1448.8)

= 0.000683 × 10^-7 m

= 68.3 × 10^-12 m = 68.3 pm

Therefore the mean De-Broglie's wavelength is 68.3 pm