The radius of circular electron orbits in the Bohr model of the hydrogen atom are given by (5.29 ✕ 10−11 m)n2, where n is the electron's energy level (see figure below). The speed of the electron in each energy level is (c/137n), where c = 3 ✕ 108 m/s is the speed of light in vacuum.
(a) What is the centripetal acceleration of an electron in the ground state (n = 1)of the Bohr hydrogen atom?
magnitude( )m/s2
direction( )
(b) What are the magnitude and direction of the centripetal force acting on an electron in the ground state?
magnitude( )N
direction( )
(c) What are the magnitude and direction of the centripetal force acting on an electron in the
n = 2 excited state?
magnitude( )N
r = (5.29 ✕ 10−11 m)n^2
c = 3*10^8 m/s
v = 3*10^8/(137 n)
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1. if n = 1
Ac = v^2/r = 9*10^18/(18769*5.29*10^-11)
= 9.06*10^-5 * 10^18 * 10^11
= 9.06 * 10^24 m/s^2 yikes!
direction is toward the proton in the center of course :)
2. force = mass of electron * Ac
Now you should be able to do #3
An airplane starts from rest and accelerates at 11.5m/s2. What is its speed at the end of a 655m runway?
I think you post your question at the wrong place. click on post a question on the top right so you can post your question for tutor to see and help you out.
To find the centripetal acceleration of an electron in the ground state of the Bohr hydrogen atom, we can use the formula for centripetal acceleration:
a = v^2 / r
where v is the velocity of the electron and r is the radius of its orbit.
In this case, the speed of the electron in the ground state is given by v = c / (137 * n), where c is the speed of light in vacuum and n is the electron's energy level. Plugging in the values, we have:
v = (3 * 10^8 m/s) / (137 * 1)
v ≈ 2.19 * 10^6 m/s
The radius of the electron orbit in the ground state is given by r = (5.29 * 10^-11 m) * (1)^2 = 5.29 * 10^-11 m.
Now, we can calculate the centripetal acceleration:
a = (2.19 * 10^6 m/s)^2 / (5.29 * 10^-11 m)
a ≈ 9.05 * 10^22 m/s^2
So, the magnitude of the centripetal acceleration of an electron in the ground state is approximately 9.05 * 10^22 m/s^2.
As for the direction, the centripetal acceleration always points towards the center of the circular path. In this case, since the electron is moving in a circular orbit, the centripetal acceleration will be directed towards the center of the orbit.
Therefore, the direction of the centripetal acceleration is towards the center of the atom.
Next, to find the centripetal force on the electron in the ground state, we can use Newton's second law:
F = m * a
where F is the force, m is the mass of the electron, and a is the centripetal acceleration.
The mass of the electron is approximately 9.11 * 10^-31 kg.
Plugging in the values, we have:
F = (9.11 * 10^-31 kg) * (9.05 * 10^22 m/s^2)
F ≈ 8.27 * 10^-8 N
So, the magnitude of the centripetal force on the electron in the ground state is approximately 8.27 * 10^-8 N.
Again, the direction of the centripetal force is towards the center of the atom, which is the same as the direction of the centripetal acceleration.
Therefore, the direction of the centripetal force is towards the center of the atom.
Finally, to find the centripetal force on an electron in the n = 2 excited state, we can follow the same steps as before.
The speed of the electron in the n = 2 excited state is v = c / (137 * 2) = (3 * 10^8 m/s) / (137 * 2) ≈ 1.09 * 10^6 m/s.
The radius of the electron orbit in the n = 2 excited state is r = (5.29 * 10^-11 m) * (2)^2 = 2.12 * 10^-10 m.
Using the same formula for centripetal acceleration, we can calculate:
a = (1.09 * 10^6 m/s)^2 / (2.12 * 10^-10 m)
a ≈ 5.56 * 10^20 m/s^2
So, the magnitude of the centripetal acceleration on the electron in the n = 2 excited state is approximately 5.56 * 10^20 m/s^2.
Once again, the direction of the centripetal acceleration is towards the center of the atom.
To find the centripetal force, we can use Newton's second law:
F = m * a
Plugging in the electron's mass and the calculated centripetal acceleration, we have:
F = (9.11 * 10^-31 kg) * (5.56 * 10^20 m/s^2)
F ≈ 5.09 * 10^-10 N
The magnitude of the centripetal force on the electron in the n = 2 excited state is approximately 5.09 * 10^-10 N.
And, once again, the direction of the centripetal force is towards the center of the atom.