Question:
... snip ...
Compute the energies of a particle under a field of constant force towards the center using Bohr's model.
Solution:
The particle is going in circular motion, so it is running on a particular plane, choose coordinate system so that the plane is the xy plane (i.e. z-coordinate is always zero), the trajectory of the particle can then be simplified to $ (r \cos \omega t, r \sin \omega t) $. The velocity is the first derivative $ (-r \omega \sin \omega t, r \omega\cos \omega t) $, and the acceleration would be the second derivative $ (-r \omega^2 \cos \omega t, -r \omega^2 \sin \omega t) $
Now let's work on the magnitudes, $ |v| = r \omega $, $ |a| = r\omega^2 $.
The total energy is given by
$ \begin{eqnarray*} E & = & \frac{p^2}{2m} + Fr \\ & = & \frac{(mv)^2}{2m} + mar \\ & = & \frac{mv^2}{2} + mar \\ & = & \frac{m(r\omega)^2}{2} + m(r\omega^2)r \\ & = & \frac{m(r\omega)^2}{2} + m(r\omega)^2 \\ & = & \frac{3m(r\omega)^2}{2} \end{eqnarray*} $
So we need to find $ r \omega $. The Bohr's model give us $ mr^2\omega = m(r\omega)r = mvr = n\hbar $, and the Newton's second law gives us $ F = ma = mr\omega^2 $. A little arithmetic trick gives
$ \begin{eqnarray*} (r\omega)^3 &=& (r^2\omega)(r\omega^2) \\ &=& \frac{mr^2\omega}{m}\frac{mr\omega^2}{m} \\ &=& \frac{n\hbar}{m}\frac{F}{m} \\ &=& \frac{n\hbar F}{m^2} \\ (r\omega) &=& (\frac{n\hbar F}{m^2})^{1/3} \end{eqnarray*} $
Substitute this back to the equation we get the answer
$ \begin{eqnarray*} E & = & \frac{3m(r\omega)^2}{2} \\ & = & \frac{3m((\frac{n\hbar F}{m^2})^{1/3})^2}{2} \\ & = & \frac{3}{2}m(\frac{n\hbar F}{m^2})^{2/3} \\ & = & \frac{3}{2}(m^{3/2}\frac{n\hbar F}{m^2})^{2/3} \\ & = & \frac{3}{2}(\frac{n\hbar F}{\sqrt{m}})^{2/3} \\ & = & \frac{3}{2}n^{2/3} (\frac{\hbar F}{\sqrt{m}})^{2/3} \end{eqnarray*} $
This formula solved question 7, 8, 9 as follow:
Question 7 requires the quantity that is a factor that is independent of $ n $, that would be $ (\frac{\hbar F}{\sqrt{m}})^{2/3} $.
Question 8 ask for the number part of the ground state, that would be $ \frac{3}{2}1^{2/3} = 1.5 $.
Question 9 ask for the exponent, so it is $ \frac{2}{3} $.
... snip ...
Compute the energies of a particle under a field of constant force towards the center using Bohr's model.
Solution:
The particle is going in circular motion, so it is running on a particular plane, choose coordinate system so that the plane is the xy plane (i.e. z-coordinate is always zero), the trajectory of the particle can then be simplified to $ (r \cos \omega t, r \sin \omega t) $. The velocity is the first derivative $ (-r \omega \sin \omega t, r \omega\cos \omega t) $, and the acceleration would be the second derivative $ (-r \omega^2 \cos \omega t, -r \omega^2 \sin \omega t) $
Now let's work on the magnitudes, $ |v| = r \omega $, $ |a| = r\omega^2 $.
The total energy is given by
$ \begin{eqnarray*} E & = & \frac{p^2}{2m} + Fr \\ & = & \frac{(mv)^2}{2m} + mar \\ & = & \frac{mv^2}{2} + mar \\ & = & \frac{m(r\omega)^2}{2} + m(r\omega^2)r \\ & = & \frac{m(r\omega)^2}{2} + m(r\omega)^2 \\ & = & \frac{3m(r\omega)^2}{2} \end{eqnarray*} $
So we need to find $ r \omega $. The Bohr's model give us $ mr^2\omega = m(r\omega)r = mvr = n\hbar $, and the Newton's second law gives us $ F = ma = mr\omega^2 $. A little arithmetic trick gives
$ \begin{eqnarray*} (r\omega)^3 &=& (r^2\omega)(r\omega^2) \\ &=& \frac{mr^2\omega}{m}\frac{mr\omega^2}{m} \\ &=& \frac{n\hbar}{m}\frac{F}{m} \\ &=& \frac{n\hbar F}{m^2} \\ (r\omega) &=& (\frac{n\hbar F}{m^2})^{1/3} \end{eqnarray*} $
Substitute this back to the equation we get the answer
$ \begin{eqnarray*} E & = & \frac{3m(r\omega)^2}{2} \\ & = & \frac{3m((\frac{n\hbar F}{m^2})^{1/3})^2}{2} \\ & = & \frac{3}{2}m(\frac{n\hbar F}{m^2})^{2/3} \\ & = & \frac{3}{2}(m^{3/2}\frac{n\hbar F}{m^2})^{2/3} \\ & = & \frac{3}{2}(\frac{n\hbar F}{\sqrt{m}})^{2/3} \\ & = & \frac{3}{2}n^{2/3} (\frac{\hbar F}{\sqrt{m}})^{2/3} \end{eqnarray*} $
This formula solved question 7, 8, 9 as follow:
Question 7 requires the quantity that is a factor that is independent of $ n $, that would be $ (\frac{\hbar F}{\sqrt{m}})^{2/3} $.
Question 8 ask for the number part of the ground state, that would be $ \frac{3}{2}1^{2/3} = 1.5 $.
Question 9 ask for the exponent, so it is $ \frac{2}{3} $.
No comments:
Post a Comment