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Chapter 38: Problem 27

An excited hydrogen atom emits a photon with an energy of \(1.133 \mathrm{eV}\). What were the initial and final states of the hydrogen atom before and after emitting the photon?

Short Answer

Expert verified
Answer: The initial state is 3, and the final state is 2.

Step by step solution

01

Recall the formula for energy levels in the hydrogen atom

The energy levels of the hydrogen atom can be determined using the following formula: \(E_n = -\dfrac{13.6 \mathrm{eV}}{n^2}\), where \(n\) is the principal quantum number (state) and \(E_n\) denotes the energy of the nth state.
02

Calculate the energy difference

The energy of the emitted photon is the difference in energy between the initial and final states of the atom. Thus, we can write the energy difference as: \(\Delta E = E_{initial} - E_{final} = 1.133 \mathrm{eV}\)
03

Use the formula for energy levels to find the possible state transitions

Since the energy difference is caused by a transition between two energy states, we can set up an equation to find the possible initial and final states: \(E_{initial} - E_{final} = \dfrac{13.6}{n^2_{final}} - \dfrac{13.6}{n^2_{initial}} = 1.133 \mathrm{eV}\)
04

Find the ratio of initial and final states

To simplify our calculation, let's call this ratio \(k\): \(k = \dfrac{1}{n^2_{initial}} - \dfrac{1}{n^2_{final}}\). Now, we can rewrite our equation as: \(13.6 \times k = 1.133 \mathrm{eV}\) Solving for \(k\), we get: \(k =\dfrac{1.133}{13.6} = 0.0833\)
05

Find the initial and final states that satisfy this ratio

Since \(n_{initial}\) and \(n_{final}\) are integers and \(n_{initial}>n_{final}\), we can find pairs of values \((n_{initial}, n_{final})\) that satisfy the condition \(k = \dfrac{1}{n^2_{initial}} - \dfrac{1}{n^2_{final}}\). After checking possible energy state transitions, we find that only the following pair of states satisfies the obtained ratio: \((n_{initial}, n_{final}) = (3, 2)\)
06

Conclude the initial and final states

Based on our calculations, the initial and final states of the hydrogen atom before and after emitting the photon are: Initial state: \(n_{initial} = 3\) Final state: \(n_{final} = 2\)

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Most popular questions from this chapter

Prove that the period of rotation of an electron on the \(n\) th Bohr orbit is given by: \(T=n^{3} /\left(2 c R_{\mathrm{H}}\right),\) with \(n=1,2,3, \ldots\)

Following the steps outlined in our treatment of the hydrogen atom, apply the Bohr model of the atom to derive an expression for a) the radius of the \(n\) th orbit, b) the speed of the electron in the \(n\) th orbit, and c) the energy levels in a hydrogen-like ionized atom of charge number \(Z\) that has lost all of its electrons except for one electron. Compare the results with the corresponding ones for the hydrogen atom.

What is the wavelength of the first visible line in the spectrum of doubly ionized lithium? Begin by writing down the formula for the energy levels of the electron in doubly ionized lithium - then consider energy-level differences that give energies in the appropriate (visible) range. Express the answer in terms of "the transition from state \(n\) to state \(n^{\prime}\) is the first visible, with wavelength \(X\)."

A hydrogen atom is in its fifth excited state, with principal quantum number \(n=6 .\) The atom emits a photon with a wavelength of \(410 \mathrm{nm}\). Determine the maximum possible orbital angular momentum of the electron after emission.

By what percentage is the electron mass changed in using the reduced mass for the hydrogen atom? What would the reduced mass be if the proton had the same mass as the electron?
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