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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.

Short Answer

Expert verified
The classical turning point, where the potential and total energy are equal for a hydrogen atom in the 1s state, is at the Bohr radius (\(r=a\)). For values greater than this (\(r>a\)), it is considered to be a classically forbidden region, yet the electron has a probability of approximately 13.53% to be found in this region.

Step by step solution

01

Understanding Potential and Total Energy

Potential energy of an electron in a hydrogen atom is given by \(U(r) = - \frac{e^2}{4Ï€\varepsilon_0 r}\) where \(e\) is the charge of the electron, \(\varepsilon_0\) is the permittivity of free space and \(r\) is the distance of the electron from the nucleus. The total energy is given by \(E = \frac{U}{2} = - \frac{e^2}{8Ï€\varepsilon_0 r}\)
02

Equating Potential Energy to Total Energy

Equating \(U(r)\) to \(E\) and simplifying, \(r = \frac{e^2}{8Ï€\varepsilon_0 E}\), we substitute the values \(e^2 = 14.4 eV. nm, E = -13.6 eV\) for the 1s state and \(\varepsilon_0 = 1\) in atomic units, hence we get, \(r = a), where \( a = 0.529 Ã… = 0.0529 nm\), this value is called the Bohr radius.
03

Determining the Classically Forbidden Region

The region for which \( U(r) > E \) is called classically forbidden. Now, since \(U(r)\) and \(E\) are already equal at \(r=a\), it implies that \(U(r) > E\) for \(r > a\), Therefore, the classically forbidden region is \(r > a\).
04

Calculating the Probability

To find the probability, integrate the radial distribution function for the 1s state from a to infinity. The radial distribution function for an electron in the 1s state of a hydrogen atom is \[P(r) = \(4\frac{(Z/a_0)^3}{a^3}\)r^2 e^{-2Zr/a_0}\] where \(Z\) is the atomic number. For hydrogen \(Z = 1\) and \(a_0 = a\), so the equation simplifies to \(P(r) = 4r^2 e^{-2r/a}\). The probability \(P\) is then \[P = ∫_a^∞ 4r^2 e^{-2r/a} dr\] Integrating by parts twice yields \(P = e^{-2}\) which is approx 13.53%.

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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 in the universe and consists of one proton and one electron. - The electron orbits the proton, creating what can be thought of as a 'mini solar system'. - This atom is fundamental to quantum mechanics because it forms the basis upon which more complex atoms are understood. Quantum mechanics governs how subatomic particles like electrons behave, which differs markedly from how larger objects behave under classical physics. In a hydrogen atom, the electron exists in specific states defined by quantum numbers that give rise to quantized energy levels. Each energy state signifies a specific energy level that the electron may occupy. The ground state or "1s state" is the lowest energy level available to an electron within the hydrogen atom and is the state of our interest here. -Since energy levels are quantized, the electron cannot exist in between these states. -The potential energy of the electron depends upon its distance from the proton, and this potential energy, along with kinetic energy, contributes to the total energy of the system.
Bohr Radius
The Bohr radius is a fundamental physical constant that represents the most probable distance between an electron and the nucleus in a hydrogen atom in its ground state. This distance is denoted by the symbol \( a \) and is approximately 0.529 Ã… (angstroms). Here's why it matters:- The Bohr radius helps define the scale of the atom and is a pioneering example of quantum mechanical applications in atomic physics.- It represents where the classical turning point occurs in the hydrogen atom - the point where the potential energy equals the total energy of the electron.At \( r = a \), or the Bohr radius, the potential energy and total energy are equal, demarking a boundary between classical and quantum behavior. This radius provides a natural unit of length when discussing atomic dimensions, with other quantum systems being compared or scaled accordingly. The understanding of such atomic scales is central in the study of atomic structures and spectra.
Quantum Tunneling
Quantum tunneling is an intriguing phenomenon where particles move through potential barriers that they seemingly shouldn't pass based on classical mechanics. In the exercise, when considering the hydrogen atom, if the electron is beyond the Bohr radius, it is in a region where classically, its potential energy exceeds its total energy, thus implying negative kinetic energy which is impossible. - Quantum mechanics permits the electron to "tunnel" through this classically forbidden area. - This ability is a direct consequence of the wave-like properties of particles as described by quantum mechanics laws. In our exercise, the calculated probability of finding an electron in this tunneling region is approximately 13.53%. This tunneling has important implications, such as allowing particles to pass where they 'shouldn't' and is critical in processes such as nuclear fusion in stars. Essentially, tunneling explains behaviors not possible under classical physics, further showcasing the unique nature of quantum mechanics.

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Most popular questions from this chapter

When our sun exhausts its nuclear fuel, it will ultimately shrink due to gravity and become a white dwarf, with a radius of approximately \(7000 \mathrm{~km}\). (a) Using the mass of the sun, \(M_{\text {sun }}=2.0 \times 10^{30} \mathrm{~kg}\). and the mass of a proton, \(M_{\text {proton }}=1.7 \times 10^{-27} \mathrm{~kg}\). estimate the number of electrons in the sun. (b) From the radius given, estimate the average volume to be occupied by each electron in the eventual white dwarf. (c) The white dwarf will consist of mostly carbon. Since there are six clectrons in each carbon atom, multiply the volume of an electron by 6 to obtain the volume of each carbon atom. (d) Model the atomic arrangement as a cubical lattice and use your earlier estimate to determine the distance \(L\) between adjacent carbon nuclei. (e) View each atom as an \(L \times L \times L\) box filled with six electrons. Since no two clectrons can be in the same quantum state, what quantum numbers \(\left(n_{X}, n_{Y}, n_{Z}, m_{z}\right)\) are associated with these six electrons, where \(m_{z}\) is the quantum number for the component of the electron spin along the z-axis? (f) Ignoring contributions from spin couplings and Coulomb interactions between the electrons, what would be the energy of the higher- energy electrons?

An electron is in the hydrogen atom with \(n=5 .\) (a) Find the possible values of \(L\) and \(L_{z}\) for this electron, in units of \(h .\) (b) For each value of \(L\), find all the possible angles between \(\overrightarrow{\boldsymbol{L}}\) and the z-axis. (c) What are the maximum and minimum values of the magnitude of the angle between \(\vec{L}\) and the \(z\) -axis?

The hydrogen spectrum includes four visible lines. Of these, the blue line corresponds to a transition from the \(n=5\) shell to the \(n=2\) shell and has a wavelength of \(434 \mathrm{nm}\). If we look closer, this line is broadened by fine structure due to spin-orbit coupling and relativistic effects. (a) How many different sets of \(l\) and \(j\) quantum numbers are there for the \(n=5\) shell and for the \(n=2\) shell? (b) How many different energy levels are there for \(n=5\) and for \(n=2 ?\) For each of these levels, what is their energy difference in eV from \(-(13.6 \mathrm{eV}) / n^{2} ?\) (c) In a transition that emits a photon the quantum number \(l\) must change by \(\pm 1 .\) Which transition in the fine structure of the hydrogen blue line emits a photon of the shortest wavelength? For this photon what is the shift in wavelength due to the fine structure? (d) Which transition in the fine structure emits a photon of the longest wavelength? For this photon what is the shift in wavelength due to the fine structure? (e) By what total extent, in \(\mathrm{nm}\), is the wavelength of the blue line broadened around the \(434 \mathrm{nm}\) value?

Show that \(\Phi(\phi)=e^{i m \phi}=\Phi(\phi+2 \pi)\) (that is, show that \(\Phi(\phi)\) is periodic with period \(2 \pi\) ) if and only if \(m_{l}\) is restricted to the values \(0,\pm 1,\pm 2, \ldots\) (Hint: Euler's formula states that \(\left.e^{j \phi}=\cos \phi+i \sin \phi .\right)\)

For each of the following states of a particle in a three-dimensional cubical box, at what points is the probability distribution function a maxi\(\mathrm{mum}\) (a) \(n_{X}=1, n_{Y}=1, n_{Z}=1\) and (b) \(n_{X}=2, n_{Y}=2, n_{Z}=1 ?\)

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