Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 36: Problem 56

A hydrogen atom is in the \(2 s\) state. Find the probability that its electron will be found (a) beyond one Bohr radius and (b) beyond 10 Bohr radii.

Short Answer

Expert verified
The probabilities of finding an electron beyond 1 and 10 Bohr radii in a hydrogen atom in the 2s state can be computed by evaluating the relevant integrals. These unsurprisingly decrease as the distance from the nucleus increases, reflecting the higher probability of finding electrons near the nucleus.

Step by step solution

01

Identify The Wavefunction

In a hydrogen atom, for the 2s state (n=2, l=0), the radial part of the wavefunction, R(r) is given as \( R(r) = \frac{1}{2\sqrt{2}} (\frac{r}{a_0}) e^{-\frac{r}{2a_0}} \) where \( a_0 \) is the Bohr radius and \( r \) is the radial distance from the nucleus of the atom.
02

Compute The Radial Probability Density

The radial probability density corresponds to the square of the absolute magnitude of the wavefunction. It represents the probability of finding an electron at a distance \( r \). The radial probability density \( |R(r)|^2 \) can therefore be computed as \( |R(r)|^2 = ( \frac{1}{2\sqrt{2}} )^2 (\frac{r}{a_0})^2 e^{-\frac{r}{a_0}} \). This should be integrated over the volume element to find the probability.
03

Calculate Probability Beyond One Bohr Radius

First, to find the probability that the electron is beyond one Bohr radius a, the volume element corresponds to the spherical shell defined by \( r \) and \( r+dr \), which is \( 4\pi r^2 dr \). The probability of finding an electron in the radial range r to r+dr, is thus given by \( P = \int_1^\infty 4\pi r^2 |R(r)|^2 dr \). This integral has to be computed numerically for the result.
04

Calculate Probability Beyond Ten Bohr Radii

Second, to find the probability that the electron is beyond ten Bohr radii, the same formula used in step 3 can be used. However, the limits of the integral will change, as the radial range of interest is now between 10 to infinity. Therefore, the integral becomes \( P = \int_{10}^\infty 4\pi r^2 |R(r)|^2 dr \). Again, this integral has to be computed numerically.

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.

Start your free trial

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

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

An infinite square well contains nine electrons. Find the energy of the highest-energy electron in terms of the ground-state energy \(\vec{E}_{1}\)

The \(4 p \rightarrow 3 s\) transition in sodium produces a double spectral line at 330.2 and \(330.3 \mathrm{nm} .\) What's the energy splitting of the \(4 p\) level?

Find the maximum possible magnitude for the orbital angular momentum of an electron in the \(n=7\) state of hydrogen.

Show that the wavelength \(\lambda\) in \(\mathrm{nm}\) of a photon with energy \(E\) in eV is \(\lambda=1240 / E\)

Is it possible for a hydrogen atom to be in the \(2 d\) state? Explain.
See all solutions

Recommended explanations on Physics Textbooks

Electric Charge Field and Potential

Read Explanation

Energy Physics

Read Explanation

Torque and Rotational Motion

Read Explanation

Nuclear Physics

Read Explanation

Rotational Dynamics

Read Explanation

Conservation of Energy and Momentum

Read Explanation
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