Consider a one-dimensional crystal (similar to a carbon nanowire) with length 10 um and lattice spacing 0.1 nm.
1 answer:
Answer:
a) Fermi level = 600 electron-volts
b)
Explanation:
Given data:
length of one-dimensional crystal = 10 um
Lattice spacing = 0.1 nm
A) Determine the Fermi level assuming one electron per atom
Total length = 10 <em>u</em>m
Interatomic separation of a = 0.1 nm
in this case the Atom has one electron therefore the number of electrons = 10^5 and the number of states Ns = gsN = 2 * 10^5 ( attached below is some part of the solution )
hence : Fermi level = 600 electron-volts
B) Determine the density of states as a function of electron energy
attached below is the detailed solution
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Answer:
Part i)
h = 5.44 m
Part ii)
h = 3.16 m
Explanation:
Part i)
Since the ball is rolling so its total kinetic energy in this case will convert into gravitational potential energy
So we have
here we know that for spherical shell and pure rolling conditions
Part b)
If ball is not rolling and just sliding over the hill then in that case
Newton's second and third law can be found the correct answer is:
- The force of the air leaving the balloon by action and reaction causes it to have a positive acceleration upward and rise.
Pressure is defined by the relationship between forces and area,
P = F / A
Where P is pressure, F is force and A is area
F = P A
Newton's second law is that the sum of the external forces of a body is proportional to the mass of the body and the acceleration
Newton's third law or action and reaction says that forces in nature appear in pairs on the two interacting bodies, the magnitude of the force is equal, but in the opposite direction; each one is applied to one of the bodies.
A free-body diagram of the process is shown in the attachment
B + F - W = m a
W = m g
Where B is the thrust, F the force of air exit, W the weight, m the mass and the acceleration of the body
Archimedes' principle states that the vertical thrust on a body is equal to the fart of the dislodged fluid, in this case air.
B =ρ_{air} g V_{air}
Where ρ and V_ {air} are the density and volume of the air, respectively,
It's substitute
ρ_{air} g V_{air} + P A - m g = m a
a =
The mass of the body is formed by the mass of the balloon plus the mass of air that it contains
Let's analyze this expression in three moments
- Before opening the hole at the bottom. The second term is zero, so the thrust is not enough to lift the balloon, the acceleration is negative.
- Just after opening. The force of the air coming out that by the law of action and reaction creates a force of equal magnitude and upward direction on the balloon. The sum of these two forces makes the acceleration positive, therefore the balloon rises
- After the pressure inside the balloon is equal to the atmospheric pressure. Tthis case the second term is zero again and the balloon begins to descend with an acceleration less than the acceleration of gravity.
In conclusion with Newton's second and third law we find the correct answer is:
- The force of the air leaving the balloon by action and reaction law causes it to have a positive acceleration upward and rise.
Learn more about Newton's law here:
