A sample of hydrogen atoms are all in the n=5 state. If all...

A sample of hydrogen atoms are all in the $n=5$ state. If all the atoms return to the ground state, how many different photon energies will be emitted, assuming all possible transitions occur? If there are 500 atoms in the sample and assuming that from any state all possible downward transitions are equally probable, what is the total number of photons that will be emitted when all of the atoms have returned to the ground state?

Key Concepts

Atomic Energy Levels

In atoms such as hydrogen, electrons occupy discrete energy levels characterized by quantum numbers. These levels arise from the quantization of energy in the atom, and each level’s energy can be calculated using formulas derived from quantum mechanics. The concept of energy levels is fundamental to understanding spectral lines and the behavior of electrons when they absorb or emit energy.

Quantum Transitions

Quantum transitions refer to the process by which electrons move between these discrete energy levels. When an electron makes a downward transition (from a higher to a lower energy level), the energy difference is released as a photon. The allowed transitions, governed by selection rules, determine which transitions can actually occur, and each transition has a unique energy difference resulting in a distinct photon energy.

Photon Emission from Level Transitions

When an atom undergoes a transition between energy levels, the energy difference between those levels is emitted as a photon. This concept connects the quantum mechanical energy structure of the atom with observable electromagnetic radiation. Different transitions produce photons of different energies, which can be identified in the emission spectrum.

Cascade Transitions

In many cases, an atom does not return directly from an excited state to the ground state but rather through a series of intermediate transitions, known as cascade transitions. Each step in the cascade emits a photon and the overall process can involve multiple photons, each corresponding to a different energy interval between the participating energy levels.

Expected Value in Stochastic Transitions

When the transition from a higher energy level can occur via several pathways, probabilistic methods are used to compute the expected number of transitions (or photons emitted) during the decay process. By assigning equal probabilities to each possible downward transition, one can use the concept of expected value from probability theory to predict the average number of photons emitted during the cascade from a given excited state to the ground state.

Transcript

00:01 In this question, we are given that hydrogen atom undergoes two downward transitions from n equals to 5 down to 2 states.

00:09 We want to find the possible combinations and the photon wavelengths emitted.

00:14 So to do this question, first we will draw out the energy levels using en equals to negative e0 over n square, which is negative, 1326.

00:30 Eb xxxx2.

00:32 So we can calculate our e2 which is negative 13 .6 divided by 4 we get negative 3 .4 eb and e3 is negative that in 6 divided by 3 square and the thing is negative 1 .5 1 eb.

00:55 Okay, i just want to move it down here.

01:06 Okay, then i continue to do e4, negative 13 .6, devout by 4 square, again negative 0 .85 ev, and then e5 is negative 13 .6, give up by 25, 5 square.

01:24 Okay, so we get negative 0 .544ev.

01:28 Okay, so we can contract the energy levels like grams.

01:31 Okay, so this will help us to look at the transitions.

01:39 Okay, so we have negative 0 .544, then negative 0 .85, negative 1, and negative 3 .4.

01:50 Okay, so this is n -quers 2, 3, 4, and 5.

02:00 Okay, so that we want to draw out two downward transitions.

02:07 So you can have n equals from 5 to 4 and then from 4 to 2.

02:13 This is one possible pathway.

02:17 You can also have going from 5 to 3 then 3 to 2.

02:21 Okay, so we have two combinations.

02:27 Okay, so combination 1.

02:33 Okay, first and equals to 5, 2 and equals to 4 and then the delta e.