Consider the following four transitions in a hydrogen atom:...

Consider the following four transitions in a hydrogen atom: (i) $n_{i}=2, n_{f}=6$ (ii) $n_{i}=2, n_{\mathrm{f}}=8$ (iii) $n_{i}=7, n_{f}=8$ (iv) $n_{i}=6, n_{f}=2$ Find (a) the longest-and (b) the shortest-wavelength photon that can be emitted or absorbed by these transitions. Give the value of the wavelength in each case. (c) For which of these transitions does the atom lose energy? Explain.

Consider the following four transitions in a hydrogen atom:
(i) $n_{i}=2, n_{f}=6$
(ii) $n_{i}=2, n_{\mathrm{f}}=8$
(iii) $n_{i}=7, n_{f}=8$
(iv) $n_{i}=6, n_{f}=2$
Find (a) the longest-and (b) the shortest-wavelength photon that can be emitted or absorbed by these transitions. Give the value of the wavelength in each case. (c) For which of these transitions does the atom lose energy? Explain.

Key Concepts

Emission versus Absorption Processes

The direction of the electron transition determines whether a photon is emitted or absorbed. If an electron moves from a higher energy level to a lower one, it emits a photon and loses energy. Conversely, a transition from a lower level to a higher level requires the absorption of a photon. This concept is crucial for understanding how atoms interact with light and change their energy state.

Rydberg Formula

For hydrogen-like atoms, the Rydberg formula provides a direct way to calculate the wavelength of emitted or absorbed light during an electron transition. It is given by 1/? = R (1/n_f² - 1/n_i²), where R is the Rydberg constant. This formula encapsulates the quantization of energy levels and is fundamental for analyzing spectral lines in hydrogen.

Energy Transition and Photon Wavelength Relation

When an electron transitions between two energy levels, the difference in energy is carried away or provided by a photon. The energy difference ?E corresponds to the photon's energy through the relation E = hc/?, where h is Planck’s constant and c is the speed of light. A larger energy difference produces a photon with a shorter wavelength, while a smaller energy difference results in a photon with a longer wavelength.

Quantized Energy Levels

In the hydrogen atom, electrons reside in discrete energy levels that are quantized. Each state is specified by a principal quantum number n, and the energy associated with a particular level is given by the formula E? = -13.6 eV/n². This quantization explains why only certain photon energies (and thus wavelengths) are observed when electrons transition between these levels.

Transcript

00:01 In this problem, we're going to talk about atomic transitions.

00:03 So let's say that we have a hydrogen atom whose energy levels, en, are equal to minus 13 .6 electron volts divided by n squared.

00:18 And let's say that the atom transitions from an initial state to a final state.

00:26 If an f is greater than an i, then the atom, gains energy and for that reason it absorbs a photon.

00:52 Now, if an f is smaller than an i, then the electron loses energy, and for that reason, it emits a photon.

01:10 And also, we need to remember that the energy of the photon is just equal to the difference of energy between the two energy levels and the energy of the photon is h .c.

01:24 Over lambda.

01:26 And we have these four transitions here.

01:30 And i is the initial energy level, and f is the final one.

01:34 And in question a, we have to find what is the longest wavelength of the photon emitted or absorbed by these transitions.

01:44 Well, since the energy of the photon is equal to h .c.

01:47 Over lambda, the longest wavelength corresponds to the smallest possible energy.

01:52 So let's calculate the energy in the transition between the two states in each case.

01:59 Just remember that the energy in its transition is equal to the energy of the final energy level minus the energy of the initial energy level.

02:14 If the energy is negative, then the item emitted a photon.

02:21 If it's positive, then the atom absorbed a fold.

02:27 And this is equal to 13 .6 electron volts times 1 over nf squared minus 1 over n i squared.

02:38 In case 1, the transition is between the state 6 and the state 2.

02:45 So delta e is equal to 13 .6 electron volts.

02:57 And actually, i'm sorry, there is one mistake here...

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