Chapter 7: Problem 57
What is an atomic orbital? How does an atomic orbital differ from an orbit?
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Chapter 7: Problem 57
What is an atomic orbital? How does an atomic orbital differ from an orbit?
These are the key concepts you need to understand to accurately answer the question.
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How is the concept of electron density used to describe the position of an electron in the quantum mechanical treatment of an atom?
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 \(n=1\)
How does de Broglie's hypothesis account for the fact that the energies of the electron in a hydrogen atom are quantized?
All molecules undergo vibrational motions. Quantum mechanical treatment shows that the vibrational energy, \(E_{\mathrm{vib}},\) of a diatomic molecule like \(\mathrm{HCl}\) is given by $$ E_{\mathrm{vib}}=\left(n+\frac{1}{2}\right) h \nu $$ where \(n\) is a quantum number given by \(n=0,1,2,\) \(3, \ldots\) and \(\nu\) is the fundamental frequency of vibration. (a) Sketch the first three vibrational energy levels for \(\mathrm{HCl}\). (b) Calculate the energy required to excite a HCl molecule from the ground level to the first excited level. The fundamental frequency of vibration for \(\mathrm{HCl}\) is \(8.66 \times 10^{13} \mathrm{~s}^{-1} .\) (c) The fact that the lowest vibrational energy in the ground level is not zero but equal to \(\frac{1}{2} h v\) means that molecules will vibrate at all temperatures, including the absolute zero. Use the Heisenberg uncertainty principle to justify this prediction. (Hint: Consider a nonvibrating molecule and predict the uncertainty in the momentum and hence the uncertainty in the position.)
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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