Calculating the frequency of an electron orbiting in a magnetic field

[John Ker]

Homework Statement

An electron in a Hydrogen atom moves around in a circular orbit of radius 0.53 x 10^-10 m. Suppose the hydrogen atom is transported into a magnetic field of 0.80 T, where the magnetic field is parallel to the orbital angular momentum.
At what frequency does the electron revolve in the magnetic field?
What is the associated speed of the electron? Assume that the radius of the orbit remains constant.

Relevant Equations

V = mv^2/R
f = v/r
f = qvb

So my thought process is as follows, since the initial centripetal force and the second magnetic force are working together, we can set up an equation to calculating final frequency.
However, I am struggling with how this can be done given so little numbers.

mvi^2/r + qvB = mv^2/r

Am I on the right track? Can someone please direct me.

Thanks!

 

[Charles Link]

My first ideas on this is that the question is not completely valid. If the electron is treated classically as a charged particle orbiting the proton=basically the Bohr atom approach, the magnetic field will have little influence on the circular orbit.
The magnetic field will result in a splitting of the "2p" states by the Zeeman effect, but the magnetic field does not influence the (1s) ground state. The subject is a somewhat advanced topic known as the Zeeman effect.
Normally free moving electrons, such as in an old television tube, will travel in circular orbits in the presence of a magnetic field. You wrote the correct equations for that case.
Perhaps somewhat else can make some sense of this question that you were presented, but I am unable to. Perhaps I need to give it further thought...

 

[kuruman]

I think this is a simple application of the planetary model for the hydrogen atom. The key assumption is that the radius of the orbit remains constant. Then one has a straightforward application of Newton's 2nd law in the centripetal direction, $F_c = m\omega^2r~$, where $F_c=\frac{ke^2}{r^2}+e\omega r B$.

 

[Charles Link]

Very good @kuruman . I would agree. This one is really a sticky problem and really requires the cleverness that you brought to the problem. Your solution is simple, but really very elusive. Good job! $\\$
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I would have taken $\Delta E=\hbar \Delta \omega=-\vec{\mu} \cdot \vec{B }$ where $\vec{\mu }=\pm \mu_B m_L$ and $m_L=0$ for the $S$ orbital and $m_L=1, 0, -1$ for the $p$ states, where $\mu_B =\frac{e \hbar}{2 m_e c}$ is the Bohr magneton in c.g.s. units. Certainly above and beyond the scope of the course that presented this problem. (with $\omega=2 \pi f$), etc. The frequency of the orbit without the magnetic field could be found from the energy in the Bohr atom model: i.e. $|E_{Bohr \, atom}|=\hbar \omega_o$. Finally $\omega=\omega_o +\Delta \omega$.
Perhaps not a perfect solution=it never will be a perfect solution when applying classical concepts to a quantum mechanical problem. $\\$
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I do like @kuruman 's solution.