Linear and anular momentum of a hydrogen electron

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SUMMARY

The discussion focuses on calculating the linear and angular momentum of a hydrogen electron in the excited state of n=4 using Bohr's theory. The radius of the orbit is determined to be 0.84668352 nm, with the linear momentum calculated as 4.9821 x 10^-25 kg*m/s and the angular momentum as 4.2182 x 10^-34 J*s. Additionally, the kinetic energy is found to be 0.850625 eV, while the potential energy is calculated as -1.70 eV, resulting in a total energy of approximately -0.851 eV.

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Homework Statement



A hydrogen is in the excited state of n=4. Using the Bohr theory of the atom, find the radius of the orbit.
Find the linear momentum of the electron. (kg*m/s)
Find the angular momentum of the electron. (J*s)

Homework Equations



r = a*n^2
? maybe L = sqrt(l(l+1)h/2*pi)

The Attempt at a Solution



I found that a for hydrogen was 5.291772*10^-11m, so I calculated that r was .84668352 nm.
After that I tried the equation for L, but it gave the wrong answer for angular momentum.

Is there any equation that can find the linear and angular momentum of the electron using the information I am given?
 
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Ok, I found the answer. Linear momentum 4.9821*10^-25 kg * m/s
Angular momentum is 4.2182*10^-34 J*s

There are also three other parts to the question:

Find its kinetic energy, potential energy and total energy. (Use eV)

I found its kinetic energy (.850625eV), but I can't find its potential energy and total energy. The kinetic energy was found by using
13.61eV/n^2
where n=4

Can anyone help me find potential and total energy?

EDIT:
Ok I found the answer now. Potential energy was found using
-e^2/(4*pi*epsilon0*r)
That gave me about -1.70 eV.

Then, I just added the kinetic and potential energy together to get about -.851 eV.
 
Last edited:

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