Kyle is flying a helicopter and is rising at 5.0m/s when he releases the bag. After 2.0s
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The De broglie wavelength of a thermal neutron at room temperature 300K = 1.5 × A°
<h3>How is the De broglie wavelength of a thermal neutron at room temperature calculated?</h3>
Temperature, T = 300K
Momentum, p = mv
Therefore v = p/m
Energy, E= 1/2 m( p/m) ²
Boltzman Energy= 3/2 KT
3/2KT = 1/2 m(p/m)²
Therefore p =
According to De broglie hypothesis, P = h ÷ λ
Therefore, λ = h ÷
= 6.6× ÷
= 0.15 ×
Therefore the De broglie wavelength of a thermal neutron at room temperature 300K = 1.5 × A°
To learn more about De broglie wavelength, refer: <u>https://brainly.in/question/6131028</u>
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