what is the maximum angular speed of the electron of a hydrogen atom in a stationary orbital?

Question

anonymous · Answer

The energy of the ground state of a hydrogen atom can be calculated easily with the bohr model ( http://en.wikipedia.org/wiki/Bohr_model), which I will leave up to you to do so. Having done so, you will find the energy of the electron in the ground state is -13.6 eV (defined from the a potential energy of zero for a free electron). This means that the electron looses 13.6 eV when it becomes bound to the atom, and the loss in PE is equivalent to a gain of Kinetic energy. Thus it will gain a kinetic energy of 13.6 eV. Converting to Joules (by multiplying by $1.602\times10^{-19}$, we get $$\frac{1}{2}mv^2=13.6\times1.602\times10^{-19}$$where $m$ is the mass of the electron. Rearranging we see that $$v=\sqrt{\frac{2\times13.6\times1.602\times10^{-19}}{9.109\times10^{-31}}}=2.187\times10^6\rm{m/s}$$

anonymous · Answer

but the question was to determine angular speed

anonymous · Answer

That easy enough to extend the answer above to angular velocity $\omega$ since $v=\omega r$, where are is the radius of the circular orbit.  Again the radius of the electron in the ground state can be determined from the Bohr model, and is equal to 0.0529 nanometers.  It can be calculated using this equation $$r=\frac{4\pi \varepsilon_0\hbar^2}{m_e e^2}$$where the variables are the same as above and $\hbar$ is Plancks reduced constant, $m_e$ is teh mass of the electron, and $e$ is the electron elementary charge.

So having got the Bohr radius, we can now solve for angular velocity from the first equation $$\omega=\frac{v}{r}=\frac{2.187\times10^6}{5.29\times1^{-11}}=4.13\times10^{16}$$
with units of per second (i.e. s$^{-1}$)

anonymous · Answer

thanks