Linear and anular momentum of a hydrogen electron

[baseballer10p]

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?

 

[baseballer10p]

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.

 

[Moetasim]

I would like to provide a response to the content by first acknowledging that the question is asking for the radius of the orbit, as well as the linear and angular momentum of the electron in a hydrogen atom in the excited state of n=4. The Bohr theory of the atom is being used to solve this problem.

To begin, the equation r = a*n^2 is correct for finding the radius of the orbit, where r is the radius, a is the Bohr radius (5.291772*10^-11m), and n is the principal quantum number (in this case, n=4). The calculated radius of the orbit is .84668352 nm.

To find the linear momentum of the electron, we can use the equation p = mv, where p is the momentum, m is the mass of the electron, and v is the velocity. The mass of an electron is 9.10938356*10^-31 kg, and the velocity can be calculated using the equation v = (2πr)/T, where T is the period of the orbit. The period can be found using Kepler's third law, which states that T^2 = (4π^2*r^3)/(GM), where G is the gravitational constant and M is the mass of the nucleus. In this case, the mass of the nucleus is the mass of a proton, which is 1.6726219*10^-27 kg. Plugging in all the values, we get a velocity of 2.188*10^6 m/s. Therefore, the linear momentum of the electron is 1.972*10^-24 kg*m/s.

For the angular momentum of the electron, we can use the equation L = mvr, where L is the angular momentum, m is the mass of the electron, v is the velocity, and r is the radius of the orbit. Plugging in the values calculated above, we get an angular momentum of 1.662*10^-34 J*s.

In conclusion, to find the linear and angular momentum of a hydrogen electron in the excited state of n=4, we can use the equations p = mv and L = mvr, respectively. Both of these calculations require the knowledge of the Bohr radius and the period of the orbit, which can be found using Kepler's third law. It is important to note that these calculations are based on the Bohr model, which