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>
#SPJ4
Iron III ion is an iron ion with a +3 charge. If iron bonded with oxygen, it would form Fe2O3 which is rust. In this case, the oxidation number on Fe is +3 and it is -2 on oxygen. This would be called iron (III) oxide.
<span>The oxidation number of an ion times the number of the ions must equal zero when added together in a molecule with no charge. In this case, there are 2 iron molecules and each has a positive 3 charge. 2x3=6. There are three oxygen molecules each with a negative 2 charge. 3x-2=-6. 6+-6=0</span>
Answer:
The angular speed of the Crab nebula pulsar is 190.3 rad/s.
Explanation:
Given that,
Time T= 33 ms = 0.033 s
The angular speed is equal to the 2π divided by time period.
We need to calculate the angular speed of the Crab nebula pulsar
Using formula of angular speed
Where, T = time
= angular speed
Put the value into the formula
Hence, The angular speed of the Crab nebula pulsar is 190.3 rad/s.
B. A sandbar is formed by water. A sand dune is formed by wind.
