/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 Consider the following energy le... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the following energy levels of a hypothetical atom: \(E_{4}\) \(-1.0 \times 10^{-19} \mathrm{~J}\) \(E_{3}\) \(--5.0 \times 10^{-19} \mathrm{~J}\) \(E_{2}\) \(--10 \times 10^{-19} \mathrm{~J}\) \(E_{1}\) \(-15 \times 10^{-19} \mathrm{~J}\) (a) What is the wavelength of the photon needed to excite an electron from \(E_{1}\) to \(E_{4} ?\) (b) What is the energy (in joules) a photon must have in order to excite an electron from \(E_{2}\) to \(E_{3} ?\) (c) When an electron drops from the \(E_{3}\) level to the \(E_{1}\) level, the atom is said to undergo emission. Calculate the wavelength of the photon emitted in this process.

Short Answer

Expert verified
The answers will be: (a) wavelength is approximately \(1.41 \times 10^{-7} \mathrm{m}\) or \(141 \mathrm{nm}\), (b) energy required is \(5 \times 10^{-19}\mathrm{J}\), and (c) wavelength of the emitted photon is approximately \(1.98 \times 10^{-7} \mathrm{m}\) or \(198 \mathrm{nm}\).

Step by step solution

01

Computation of wavelength for the photon to excite an electron from \(E_{1}\) to \(E_{4}\)

Calculate the energy difference (\(ΔE\)) between these two energy levels. \(ΔE = E_{4} - E_{1} = -1.0 \times 10^{-19} \mathrm{~J} - (-15 \times 10^{-19}\mathrm{~J}) = 14 \times 10^{-19}\mathrm{~J}\). Use Planck's equation to find the wavelength, \(λ = \frac{hc}{ΔE}\) where \(h = 6.62607015 \times 10^{-34} \mathrm{~m^2 kg / s}\) (Planck's constant) and \(c = 3.0 \times 10^8 \mathrm{~m/s}\) (speed of light). Substitute these values into the equation to get the wavelength.
02

Computation of photon's energy to excite an electron from \(E_{2}\) to \(E_{3}\)

Calculate the energy difference (\(ΔE\)) between these two levels. \(ΔE = E_{3} - E_{2} = -5.0 \times 10^{-19} \mathrm{~J} - (-10 \times 10^{-19}\mathrm{~J}) = 5 \times 10^{-19}\mathrm{~J}\). This is the energy required to excite an electron from \(E_{2}\) to \(E_{3}\), thus no further calculations are needed.
03

Computation of wavelength for the emitted photon when an electron drops from \(E_{3}\) level to \(E_{1}\) level

Calculate the energy difference (\(ΔE\)) between these two levels. \(ΔE = E_{3} - E_{1} = -5.0 \times 10^{-19} \mathrm{~J} - (-15 \times 10^{-19}\mathrm{~J}) = 10 \times 10^{-19}\mathrm{~J}\). Use Planck's equation to find the wavelength, \(λ = \frac{hc}{ΔE}\) where \(h\) and \(c\) have the same values as in step 1. Substitute these values into the equation to get the wavelength.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Planck's Equation
At the heart of quantum mechanics is Planck's equation, a fundamental formula that relates the energy of a photon to its frequency. The equation is represented as \(E = hu\), where \(E\) is the energy of the photon, \(h\) is Planck's constant (\(6.62607015 \times 10^{-34} \mathrm{m^2 kg / s}\)), and \(u\) is the frequency of the photon.

In the context of atomic energy levels, when an electron moves between different energy states, a photon is either absorbed or emitted, and its energy exactly matches the difference in energy between these states. This is pivotal, as it explains how quantized energy levels within atoms result in the absorption or emission of light at specific wavelengths. The elegance of Planck's equation is in its ability to describe this very discrete interaction between light and matter that classical physics could not.
Photon Wavelength Calculation
The calculation of a photon's wavelength involves the rearrangement of Planck's equation into \(\lambda = \frac{hc}{\Delta E}\), linking the photon's wavelength \(\lambda\) with the energy change \(\Delta E\) during an electron's transition between energy levels.

In any photon wavelength calculation, we must be aware of the direct inverse relationship between wavelength and energy. This means that higher energy transitions result in shorter wavelengths, which correspond to light toward the violet end of the visible spectrum, whereas lower energy transitions result in longer wavelengths, which are toward the red end. The speed of light (\(c\)) and Planck's constant (\(h\)) are the constants that enable us to bridge the gap between theoretical energy changes within the atom and the practical wavelength of light that we can measure.
Excitation of Electrons
The excitation of electrons is a fundamental process where electrons absorb energy and move to higher energy levels within an atom. Understanding this process is critical for interpreting how atoms interact with light, leading to phenomena such as spectroscopy and lasers.

When a photon with the correct amount of energy corresponding to the energy difference between two levels strikes an atom, it can be absorbed, causing an electron to make this leap. If the photon's energy is too low or too high, the electron will not be excited, which is why only specific wavelengths of light are absorbed by an atom. This selective absorption is what gives rise to the line spectra, which provides a unique fingerprint for each element.
Emission of Photons
Conversely, the emission of photons occurs when an electron in a higher energy state relaxes to a lower energy state, releasing the excess energy in the form of a photon of light. This is essentially the reverse of excitation.

When an electron returns to a lower energy level, conservation of energy dictates that the energy it loses must go somewhere, and in the context of atomic transitions, it is emitted as a photon. The color (or wavelength) of the emitted photon corresponds to the difference in energy between the two levels involved in the transition. This process is integral to our understanding of phenomena such as the glow of neon lights and the emission spectra that scientists use to determine the composition of distant stars.

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Most popular questions from this chapter

How is the concept of electron density used to describe the position of an electron in the quantum mechanical treatment of an atom?

In the beginning of the twentieth century, some scientists thought that a nucleus may contain both electrons and protons. Use the Heisenberg uncertainty principle to show that an electron cannot be confined within a nucleus. Repeat the calculation for a proton. Comment on your results. Assume the radius of a nucleus to be \(1.0 \times 10^{-15} \mathrm{~m} .\) The masses of an electron and a proton are \(9.109 \times 10^{-31} \mathrm{~kg}\) and \(1.673 \times 10^{-27} \mathrm{~kg},\) respectively. (Hint: Treat the diameter of the nucleus as the uncertainty in position.)

An electron in a hydrogen atom is excited from the ground state to the \(n=4\) state. Comment on the correctness of the following statements (true or false). (a) \(n=4\) is the first excited state. (b) It takes more energy to ionize (remove) the electron from \(n=4\) than from the ground state. (c) The electron is farther from the nucleus (on average) in \(n=4\) than in the ground state. (d) The wavelength of light emitted when the electron drops from \(n=4\) to \(n=1\) is longer than that from \(n=4\) to \(n=2\). (e) The wavelength the atom absorbs in going from \(n=1\) to \(n=4\) is the same as that emitted as it goes from \(n=4\) to \(n=1\)

Thermal neutrons move at speeds comparable to those of air molecules at room temperature. These neutrons are most effective in initiating a nuclear chain reaction among \({ }^{235} \mathrm{U}\) isotopes. Calculate the wavelength (in \(\mathrm{nm}\) ) associated with a beam of neutrons moving at \(7.00 \times 10^{2} \mathrm{~m} / \mathrm{s}\). (Mass of a neutron \(\left.=1.675 \times 10^{-27} \mathrm{~kg} .\right)\)

In an electron microscope, electrons are accelerated by passing them through a voltage difference. The kinetic energy thus acquired by the electrons is equal to the voltage times the charge on the electron. Thus, a voltage difference of \(1 \mathrm{~V}\) imparts a kinetic energy of \(1.602 \times 10^{-19} \mathrm{C} \times \mathrm{V}\) or \(1.602 \times\) \(10^{-19} \mathrm{~J}\). Calculate the wavelength associated with electrons accelerated by \(5.00 \times 10^{3} \mathrm{~V}\).

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