Consider a Hydrogen atom with the electron in the n 8 shell. What is the energy of this system? (The magnitude of the ground sta
1 answer:
Answer:
The energy of an electron in the 8th shell is given by: -0.2125 eV
The number of subshells is: 8
The number of orbitals is: 64
The number of electrons that fit on this shell is: 128
Explanation:
First, we find the energy of the electrons in the 8th shell. In order to do this, we recall that the energy of an electron (in the Hydrogen atom) whose principal number is n is given by:
Substituting n=8, we find that the energy is given by:
In order to find the number of subshells we recall that, for a given principal quantum number n, the possible values of the quantum number l, which corresponds to the number of subshells are:
0, 1, 2, ... , n-1
Since n = 8 in our problem, the possible values of l are: 0, 1, 2, 3, 4, 5, 6, 7. Therefore, the number of subshells are 8.
Now we continue with the number of orbitals. For every subshell l, we have 2l+1 possible values of m, which correspond to the orbitals. Since the possible values of l are: 0,1,2,3,4,5,6,7, therefore, we have to perform the sum:
And we can conclude that the number of orbitals is equal to 64.
Finally, we know that we can fit two electrons per orbital, therefore we can have 64*2 = 128 electrons in the shell corresponding to n=8.
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Answer: 2.1 × 10^7 m/s
Explanation:
Please see the attachments below
Answer:
Acceleration is -30000 mi/h²
Distance travelled in the 3 seconds of deceleration is 261.888 feet
Explanation:
t = Time taken for the car to slow down = 3 s =
u = Initial velocity = 75 mi/h
v = Final velocity = 50 mi/h
s = Displacement
a = Acceleration
Equation of motion
Acceleration is -30000 mi/h²
Converting to feet
1 mile = 5280 feet
0.0496 mile = 0.0496×5280 = 261.888 feet
Distance travelled in the 3 seconds of deceleration is 261.888 feet
A) The electric field inside the paint layer is zero
B) The electric field just outside the paint layer is (radially inward)
C) The electric field at 6.00 cm from the surface is (radially inward)
Explanation:
A)
We can solve the problem by applying Gauss Law, which states that the electric flux through a Gaussian surface must be equal to the charge contained in the surface divided by the vacuum permittivity:
where
E is the magnitude of the electric field
dS is the element of the surface
q is the charge contained within the surface
is the vacuum permittivity
By taking a sphere centered in the origin,
where is the surface of the Gaussian sphere of radius r.
In this problem, we want to find the electric field just inside the paint layer, so we take a value of r smaller than
(radius of the plastic sphere is half of the diameter)
Since the charge is all distributed over the plastic sphere, the charge contained within the Gaussian sphere is zero:
And therefore,
So, the electric field inside the plastic sphere is zero.
B)
Here we apply again Gauss Law:
In this case, we want to calculate the electric field just outside the paint layer: this means that we take r as the radius of the plastic sphere, so
The charge contained within the Gaussian sphere is therefore
Therefore, the electric field is
And the negative sign indicates that the direction of the field is radially inward (because the charge that generates the field is negative). However, the text of the question says "Enter positive value if the field is directed radially inward and negative value if the field is directed radially outward", so the answer to this part is
C)
For this part again, we apply Gauss Law:
In this case, we want to calculate the field at a point 6.00 cm outside the surface of the paint layer; this means that the radius of the Gaussian sphere must be
r = 9 cm + 6 cm = 15 cm = 0.15 m
While the charge contained within the sphere is again
Therefore, the electric field in this case is
And again, this is radially inward, so according to the sign convention asked in the problem,
Learn more about electric fields:
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