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A student of classical physics says, "A charged particle. like an electron orbiting in u simple atom. shouldn't have only certain stable energies: in fact, it should lose energy by electromagnetic radiation until the atom collapses." Answer these two complaints qualitatively. appealing to as few fundumentul claims of quantum mechanics as possible

Short Answer

Expert verified
According to Quantum Mechanics, an electron in an atom can exist only at certain energy states and does not continuously lose energy via radiation. When an electron transitions from a higher state to a lower energy state, it emits a photon whose energy corresponds to the energy difference. The energy conservation prevents the atom from collapsing.

Step by step solution

01

Understanding Quantum Behaviors

Quantum Mechanics introduces the concept of 'Quantization', which forms a cornerstone of our understanding of microscopic systems. The concept of quantization suggests that energy levels are discrete, or 'quantized', which means energy can only exist in certain defined, set amounts.
02

Addressing the Energy States

For a charged particle in an atom, such as an electron, it can't just take on any energy value. It must reside in one of the particular energy levels. These energy levels correspond to the stable orbits in which electrons can reside within the atom. The energy of an electron isn't arbitrary because if it were, the negatively charged electron would be drawn into the positive nucleus, causing what is often called an 'atomic collapse'.
03

Discussing Energy loss by Electromagnetic Radiation

Now addressing the second complaint, according to classical physics, a charged particle like an electron moving in a circular orbit should emit electromagnetic radiation. This would cause it to lose energy and spiral into the nucleus. However, in quantum mechanics, an electron in a stable (i.e., non-excited) state does not emit radiation. Only when it transitions from a higher energy level to a lower one does it emit a photon, while the energy of the photon corresponds to the difference in energy levels. Additionally, because energy levels are quantized, not just any transition (and therefore not just any photon/energy) is possible, preventing a continuous energy drain of the electron.

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Key Concepts

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

Energy Quantization
In the realm of quantum mechanics, energy quantization is a fascinating concept. It's all about how energy, at the microscopic level, doesn't just vary smoothly. Instead, energy exists in specific, fixed units called 'quanta'. Think of it as stepping up rungs on a ladder, where you cannot stand between rungs. Similarly, electrons in an atom can only occupy certain energy levels or 'rungs', if you will.

This stands in contrast to what you might expect from classical physics, where energy levels are continuous and not limited to discrete values. The idea of quantization helps us understand why certain atoms are stable and don't just lose energy randomly. It also explains a great deal about how atoms interact with light and other forms of electromagnetic radiation.
  • Energy comes in fixed packets (quanta)
  • Electrons can only exist in defined energy levels
  • Prevents arbitrary loss of energy
Atomic Stability
The stability of atoms is a fundamental concept in quantum mechanics. Classical physics might suggest that an electron would gradually lose energy, eventually spiraling into the nucleus and leading to an atomic collapse. However, thanks to quantum mechanics, we know this isn't the case.

Electrons inhabit specific energy levels and cannot exist in between these levels. This quantization implies that electrons remain in specific, stable orbits unless an external force acts to move them to a different energy level. This stability guarantees that electrons do not just plummet into the nucleus despite their attraction to it.
  • Electrons have fixed positions in energy levels
  • Prevents electrons from losing energy continuously
  • Leads to the enduring stability of atoms
Electromagnetic Radiation
Classical physics would predict that an orbiting electron would inevitably emit electromagnetic radiation continuously, losing energy and eventually crashing into the nucleus. However, quantum mechanics offers a much more apt description.

In quantum mechanics, electromagnetic radiation is only emitted when an electron transitions from a higher energy level to a lower energy level. Rather than emitting radiation constantly, electrons only emit focused bursts of energy, known as photons. These photons correspond to very specific energy changes, ensuring that energy is not continuously lost. This means that not every slight movement or change of an electron results in emission. Only certain transitions, governed by quantized energy levels, produce radiation.
  • Electrons emit energy only during specific transitions
  • No constant radiation emitted from stable states
  • Ensures conservation of energy in atoms

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Most popular questions from this chapter

A study of classical waves tells us that a standing wave can be expressed as a sum of two traveling waves. Quantum-mechanical traveling waves, discussed in Chapter \(4,\) are of the form \(\Psi(x, t)=A e^{i(h x-\alpha t)} .\) Show that the infinite well's standing-wave function can be expressed as a sum of two traveling waves.

Exercises \(90-92\) refer to a particle described by the wave function $$ \psi(x)=\sqrt{\frac{2}{\pi}} a^{3 / 2} \frac{1}{x^{2}+a^{2}} $$ Show that the normalization constant is correct.

Consider the differential equation \(d^{2} f(x) / d x^{2}=b f(x)\). (a) Suppose that \(f_{1}(x)\) and \(f_{2}(x)\) are solutions. That is, $$ \frac{d^{2} f_{1}(x)}{d x^{2}}=b f_{1}(x) \text { and } \frac{d^{2} f_{2}(x)}{d x^{2}}=b f_{2}(x) $$ Show that the equation also holds when the linear combination \(A_{1} f_{1}(x)+A_{2} f_{2}(x)\) is inserted. (b) Suppose that \(f_{3}(x)\) and \(f_{4}(x)\) are solutions of \(d^{2} f(x) / d x^{2}=b f^{2}(x)\). Is \(A_{3} f_{3}(x)+A_{4} f_{4}(x)\) a solution? Justify your answer.

Whereas an infinite well has an infinite number of bound states, a finite well does not. By relating the well height \(U_{0}\) to the kinetic energy and the kinetic energy (through \(\lambda\) ) to \(n\) and \(L\). show that the number of bound states is given roughly by \(\sqrt{8 m L^{2} U_{0} / h^{2}}\). (Assume that the number is Large.)

Exercises \(94-97\) refer to a bound particle of mass \(m\) described by the wave function $$ \psi(x)=A x e^{-x^{2} / 2 b^{2}} $$ Given that the particle's total energy \(E\) is 0 , show that the potential energy is \(U(x)=\frac{h^{2}}{2 m b^{4}} x^{2}-\frac{3 h^{2}}{2 m b^{2}}\)

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