/*! 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 51 The normalized radial wave funct... [FREE SOLUTION] | 91Ó°ÊÓ

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The normalized radial wave function for the \(2 p\) state of the hydrogen atom is \(R_{2 p}=\left(1 / \sqrt{24 a^{5}}\right) r e^{-r / 2 a} .\) After we average over the angular variables, the radial probability function becomes \(P(r) d r=\left(R_{2 p}\right)^{2} r^{2} d r .\) At what value of \(r\) is \(P(r)\) for the \(2 p\) state a maximum? Compare your results to the radius of the \(n=2\) state in the Bohr model.

Short Answer

Expert verified
The value of \(r\) for which the radial probability function \(P(r)\) for the \(2p\) state of a hydrogen atom is a maximum is \(a/4\). Compared to the Bohr model for the \(n=2\) state, in which the radius is \(4a_0\), the maximum of the radial distribution function is less than the radius of the Bohr model.

Step by step solution

01

Formulate the Radial Probability Function

Substitute the given radial wave function into the radial probability function to get: \(P(r) = \left[\left(\frac{1}{\sqrt{24 a^{5}}}\right) r e^{-r / 2 a}\right]^{2} r^{2}\)
02

Simplify the Function

Simplify \(P(r)\) to a form suitable for differentiation: \(P(r) = \frac{r^4}{24 a^5}e^{-r/a}\)
03

Take the Derivative

Take the derivative of \(P(r)\) with respect to \(r\), and equate it to \(0\) to find extrema: \(P'(r) = \frac{d}{dr} \left[\frac{r^4}{24 a^5}e^{-r/a}\right] = 0\) . The derivative of a product can be written as \(u'v + uv'\), where \(u\) and \(v\) are functions of \(r\). Hence, \(P'(r) = \frac{4 r^3}{24 a^5}e^{-r/a} - \frac{r^4}{24 a^6}e^{-r/a} = 0\)
04

Solve for \(r\)

Rearranging the equation gives, \(r = \frac{a}{4}\) which gives the maximum point of the distribution.
05

Compare to Bohr Model

The radius of the \(n=2\) state in the Bohr model is \(4a_0\), where \(a_0\) is the Bohr radius. Hence, we see that the peak of the radial distribution function is at \(a/4\) and not at the most probable radius for the Bohr model.

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Key Concepts

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

Hydrogen Atom
The hydrogen atom is the simplest atom, made up of just one proton and one electron. It serves as the foundational model for understanding atomic structure and quantum mechanics. In essence, this basic structure is vital for our comprehension of more complex atoms.
  • In hydrogen, the single electron orbits the nucleus, which is simply a proton.
  • The hydrogen atom is used extensively in physics as a model to explore and illustrate quantum mechanics principles.
  • Due to its simplicity, the hydrogen atom is analytically solvable, providing insights into the behavior of electrons in more complex systems.
To mathematically describe the behavior of the electron in a hydrogen atom, physicists use wave functions. These functions help to determine the probability of finding the electron in various positions around the nucleus. Understanding these probability distributions, such as through the radial wave function, is crucial in quantum mechanics.
Radial Wave Function
Radial wave functions are a component of the wave functions used in quantum mechanics to describe the behavior of electrons in atoms.
They specifically handle variations in probability in relation to the distance from the nucleus.For hydrogen-like atoms, the overall wave function \(\psi(r, \theta, \phi)\) can be separated into radial (r) and angular parts (\(\theta\) and \(\phi\)), where \(R(n, l, r)\) represents the radial part.
  • In the case of the hydrogen atom, the radial wave function provides information about how the probability of locating an electron varies with the radius from the nucleus, which is crucial for understanding electron arrangements.
  • The radial wave function for different states enables us to determine where the electron is likely to be found at a given energy level.
In this exercise, the radial wave function for the 2p state was given as \(R_{2 p}=\left(1 / \sqrt{24 a^{5}}\right) r e^{-r / 2 a}\). Maximizing the radial probability function, which is derived from the radial wave function, helps identify the most probable distance from the nucleus where the electron can be found. Understanding how to deal with these functions contributes significantly to grasping quantum behavior in atoms.
Bohr Model
The Bohr Model is a classical model that describes the hydrogen atom. It was introduced by Niels Bohr in 1913 and was a major advancement in understanding atomic structure at the time.
  • The model suggests that electrons orbit the nucleus in fixed, circular paths or "shells" without radiating energy until moving between these orbits.
  • In the Bohr model, each orbit corresponds to a specific energy level.
  • For hydrogen, the radius of the electron’s orbit at the nth energy level is given by \(r_n = n^2 a_0\), where \(a_0\) is the Bohr radius.
The Bohr model provides a basic understanding of quantum jumps—a process where the electron can move between different energy levels by absorbing or emitting energy in the form of photons. Although simplified, the model proved vital in establishing early quantum theory grounds. In the exercise, the peak of the radial distribution function for the 2p state is at \(a/4\), which differs from the Bohr model's prediction of \(4a_0\) for the n=2 state. This highlights the differences between quantum mechanics and classical physics, illustrating the intricate and often non-intuitive nature of quantum systems.

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Most popular questions from this chapter

BIO Atoms of Unusual Size. In photosynthesis in plants, light is absorbed in light-harvesting complexes that consist of protein and pigment molecules. The absorbed energy is then transported to a specialized complex called the reaction center. Quantum-mechanical effects may play an important role in this energy transfer. In a recent experiment, researchers cooled rubidium atoms to a very low temperature to study a similar energy-transfer process in the lab. Laser light was used to excite an electron in each atom to a state with large \(n .\) This highly excited electron behaves much like the single electron in a hydrogen atom, with an effective (screened) atomic number \(Z_{\text {cff }}=1\). Because \(n\) is so large, though, the excited electron is quite far from the atomic nucleus, with an orbital radius of approximately \(1 \mu \mathrm{m},\) and is weakly bound. Using these so-called Rydberg atoms, the researchers were able to study the way energy is transported from one atom to the next. This process may be a model for understanding energy transport in photosynthesis. (Source: "Observing the Dynamics of Dipole-Mediated Energy Transport by Interaction Enhanced Imaging." by G. Gunter et al., Science 342(6161) \(954-956,\) Nov. \(2013 .)\) In the Bohr model, what is the principal quantum number \(n\) at which the excited electron is at a radius of \(1 \mu \mathrm{m} ?\) (a) \(140 ;\) (b) 400 ; (c) \(20 ;\) (d) 81 .

Effective Magnetic Field. An clectron in a hydrogen atom is in the \(2 p\) state. In a simple model of the atom, assume that the electron circles the proton in an orbit with radius \(r\) equal to the Bohr-model radius for \(n=2 .\) Assume that the speed \(v\) of the orbiting electron can be calculated by sctting \(L=m v r\) and taking \(L\) to have the quantum-mechanical value for a \(2 p\) state. In the frame of the electron, the proton orbits with radius \(r\) and speed \(v .\) Model the orbiting proton as a circular current loop. and calculate the magnetic field it produces at the location of the clectron.

(a) The doubly charged ion \(\mathrm{N}^{2+}\) is formed by removing two electrons from a nitrogen atom. What is the ground-state electron configuration for the \(\mathrm{N}^{2+}\) ion? (b) Estimate the energy of the least strongly bound level in the \(L\) shell of \(\mathrm{N}^{2+}\). (c) The doubly charged ion \(\mathrm{P}^{2+}\) is formed by removing two electrons from a phosphorus atom. What is the ground-state electron configuration for the \(\mathrm{P}^{2+}\) ion? (d) Estimate the energy of the least strongly bound level in the \(M\) shell of \(\mathrm{P}^{2+}\)

Consider a hydrogen atom in the \(1 s\) state. (a) For what value of \(r\) is the potential energy \(U(r)\) cqual to the total energy \(E\) ? Express your answer in terms of \(a\). This value of \(r\) is called the classical turning point, since this is where a Newtonian particle would stop its motion and reverse direction. (b) For \(r\) greater than the classical turning point, \(U(r)>E\). Classically, the particle cannot be in this region, since the kinetic energy cannot be negative. Calculate the probability of the electron being found in this classically forbidden region.

Show that the total number of atomic states (including different spin states) in a shell of principal quantum number \(n\) is \(2 n^{2}\). [Hint: The sum of the first \(N\) integers \(1+2+3+\cdots+N\) is equal to \(N(N+1) / 2 .]\) (b) Which shell has 50 states?

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