Question

2.
What is de-Broglie wavelength of a He-atom in a container at 300 K. (Use Uavg)
(A) 0.19 Å
(B) 0.29 Å
(C) 0.79 Å
(D) 0.59 Å​

Answer 1

Answer:

0.29 Å your answer

Explanation:

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Answer 2

Explanation:

de-Broglie equation relates the wave-particle duality of matter as:

λ= h/mUavg

Average velocity can be calculated by using the equation:

Uavg = √(8RT)/πM

Given parameters are:

M=4×10^−3 kg, R=8.314 JK −1 mol −1

T=300 K, π =3.14

Average velocity =1260.42 m/s.

And by using mass of He atom, m = M/N= 4×10^

−3/6.023×10^23 kg

Planck's constant, h=6.63×10^−34 Js

de-broglie wavelength, λ=0.79 A°