Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The hydrogen atom wave function \(\psi_{200}\) is zero when \(r=2 a_{0} .\) Does this mean that the electron in that state can never be observed at a distance of \(2 a_{0}\) from the nucleus or that the electron can never be observed passing through the spherical surface defined by \(r=2 a_{0}\) ? Is there a difference between those two descriptions?

Short Answer

Expert verified
Answer: The electron cannot be observed at a distance of \(2a_0\) from the nucleus in the state \(\psi_{200}\) as the probability density at that specific position is zero. However, this does not imply that the electron can never be observed passing through the spherical surface defined by \(r=2a_0\). The electron can still transition from one region to another via different paths, which might include passing through regions very close to, but not exactly at, the spherical shell \(r=2a_0\).

Step by step solution

01

Understanding the wave function

The wave function of an electron, in this case, \(\psi_{200}\), represents the probability amplitude for an electron to be found at a certain point in space. The square of the magnitude of the wave function, \(|\psi_{200}|^2\), gives the probability density of finding the electron at that position.
02

Analyzing the given wave function at position \(r=2a_0\)

We are given that the wave function at \(r=2a_0\) is zero, i.e., \(\psi_{200}(2a_0)=0\). Since \(|\psi_{200}(2a_0)|^2=0\), it implies that the probability density of finding the electron at exactly \(r=2a_0\) is zero.
03

Differentiating the two descriptions

The first description, stating that "the electron in that state can never be observed at a distance of \(2a_0\) from the nucleus", means that the electron cannot be found at the exact position \(r=2a_0\) in space. The second description, stating that "the electron can never be observed passing through the spherical surface defined by \(r=2a_0\)", means that the electron cannot be found in a region surrounding the point \(r=2a_0\) while transitioning from one region to another.
04

Deducing the correct description

Based on our analysis, we can conclude that the first description is accurate in this case. Since the probability density of finding the electron at exactly \(r=2a_0\) is zero, the electron cannot be observed at a distance of \(2a_0\) from the nucleus in state \(\psi_{200}\). As for the second description, it is not true that the electron can never be observed passing through the spherical surface defined by \(r=2a_0\). Even though it cannot be found exactly at \(r=2a_0\), it can still transition from one region to another via different spatial paths, which might include passing through regions very close to, but not exactly at, the spherical shell \(r=2a_0\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Probability Density
Probability density is a crucial concept in quantum mechanics. It helps us understand where an electron might be found in space. When dealing with wave functions, like the hydrogen atom's \(\psi_{200}\), the square of the wave function's magnitude \(\left|\psi_{200}\right|^2\) gives us the probability density.
This quantity indicates the likelihood of finding an electron in a particular location. In our problem, \(\psi_{200}(2a_0) = 0\), which means that at \(r = 2a_0\), the probability density is zero.
This tells us there is absolutely no chance of finding the electron at the exact distance \(r = 2a_0\) from the nucleus.

However, it’s important to note:
  • A zero probability density at a specific point doesn't mean the electron can't pass nearby or go through a region close to it.
  • Probability density helps visualize potential areas where electrons are likely found but doesn't track their paths.
Hydrogen Atom
The hydrogen atom is often used as a fundamental example in quantum mechanics due to its simplicity. It consists of a single proton nucleus and a single electron. Despite this simplicity, understanding the hydrogen atom's electron behavior requires quantum mechanics.
The shape and size of the electron's 'cloud' around the nucleus can be described using solutions to the Schrödinger equation for the hydrogen atom.

Key aspects of hydrogen atom wave functions include:
  • Each wave function, such as \(\psi_{200}\), corresponds to specific energy levels and quantum states.
  • The quantum numbers \(n, l, m\) dictate the shape and orientation of these solutions, influencing where the electron is likely to be found.
  • The hydrogen atom model provides foundational insights, helpful in understanding more complex atoms.
Quantum Mechanics
Quantum mechanics is the branch of physics that studies the behavior of matter and energy at very small scales, like atoms and subatomic particles. It introduces a fundamental shift from classical physics by suggesting that particles don’t have definite positions until they're observed.
Wave functions are one of its core concepts, allowing us to calculate probabilities instead of certainties about a particle's location or state.

Features of quantum mechanics in context:
  • It uses complex mathematical equations to describe nature's smallest scales.
  • Introduces concepts such as superposition and entanglement, where particles can exist in multiple states at once and influence each other instantaneously over distance, respectively.
  • Quantum mechanics challenges our everyday experience of how the world works, offering explanations for the behavior of electrons, like that within a hydrogen atom.
  • It helps predict and explain observations that classical physics cannot, making it a cornerstone of modern physics understanding.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free