Calculating Mean Lifetime and Revolutions in Excited State of Hydrogen Atom

[stunner5000pt]

UNLIke most of the people who keep bumping their thread up by posting in it, i will bump this post up like that and SHAME ON YOU FOR NOT POSTING PART SOLUTION FOR YOUR EFFORT

but on to the question i have

The mean lifetime of the hydrogen atom in the first excited state is 10ns

What is the mean number of time an electron in the first excited state will revolve about the nucleus before returning to the ground sate with emission of a photon?

first of all n = 2, first EXCITED state

now i know \Delta E \Delta t \geq \frac{h}{2 \pi}

i can calculate the energy of the electron in this state but the mean number of times?? Is there something in the statistics that I am not geting here?

 

[Doc Al]

Hint: Use the Bohr model of the atom to figure out the time it takes for one revolution.

 

[wais]

Thank you for bringing attention to the importance of posting solutions to problems and not just bumping threads. It is important for the community to share knowledge and help each other learn, rather than just seeking answers without putting in effort.

To calculate the mean number of times an electron in the first excited state will revolve about the nucleus before returning to the ground state, we need to use the concept of the quantum mechanical model of the hydrogen atom. According to this model, the energy of the electron in the first excited state is given by:

E = -13.6 eV * \frac{1}{2^2} = -3.4 eV

Using the equation \Delta E \Delta t \geq \frac{h}{2 \pi}, we can calculate the minimum time it takes for the electron to transition from the first excited state to the ground state. Plugging in the values, we get:

\Delta t \geq \frac{2 \pi}{h} * \frac{1}{3.4 eV} = 4.6 * 10^{-17} seconds

Now, the mean lifetime of the hydrogen atom in the first excited state is given to be 10ns, which is much larger than the minimum time calculated above. This means that the electron can go through multiple transitions before emitting a photon and returning to the ground state.

To calculate the mean number of revolutions, we can divide the mean lifetime by the minimum time calculated above:

Mean number of revolutions = \frac{10 ns}{4.6 * 10^{-17} seconds} = 2.2 * 10^{16} revolutions

This means that on average, the electron in the first excited state will revolve about the nucleus 2.2 * 10^{16} times before returning to the ground state with emission of a photon. This number may seem large, but it is a result of the quantum nature of the hydrogen atom. I hope this helps clarify your doubts.