Newton's 2nd Law Applied to a Particle In Uniform Circular Motion

[Faiza]

Q. In the Bohr model of the hydrogen atom, the speed of the electron is
approximately 2.20 x 10^6 m/s. Find (a) the force acting on the electron
as it revolves in a circular orbit of radius 0.530 x 10^-10 m and (b) the
centripetal acceleration of the electron.

I'm going to post my answer in a minute. Let me know if anyone can help me figure it out

 

[Pyrrhus]

Faiza, review that topic, this is a simple plug into equation problem.

 

[M98Ranger]

Sure, I'd be happy to help!

To find the force acting on the electron, we can use Newton's second law, which states that force is equal to mass multiplied by acceleration (F=ma). In this case, the mass of the electron is 9.11 x 10^-31 kg and the acceleration is the centripetal acceleration, which we can calculate using the formula a=v^2/r, where v is the speed and r is the radius. Plugging in the values, we get:

a= (2.20 x 10^6 m/s)^2 / 0.530 x 10^-10 m = 8.73 x 10^20 m/s^2

Now, we can plug this value into the formula for force:

F = (9.11 x 10^-31 kg) * (8.73 x 10^20 m/s^2) = 7.95 x 10^-10 N

Therefore, the force acting on the electron is approximately 7.95 x 10^-10 Newtons.

To find the centripetal acceleration, we can use the same formula as before, but this time we will use the given speed of the electron and the radius of the orbit:

a= (2.20 x 10^6 m/s)^2 / 0.530 x 10^-10 m = 8.73 x 10^20 m/s^2

This is the same value we calculated earlier for the acceleration. So, the centripetal acceleration of the electron is 8.73 x 10^20 m/s^2.

I hope this helps! Let me know if you have any other questions or need further clarification.