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A hydrogen atom in an excited bound state labeled with principal quantum number \(n=3\) absorbs a photon that has wavelength \(\lambda\). The atom is ionized and the electron has kinetic energy \(8.00 \mathrm{eV}\) after it has left the atom. What was the wavelength \(\lambda\) of the photon?

Short Answer

Expert verified
The wavelength of the photon is approximately 130 nm.

Step by step solution

01

Calculate the initial energy of hydrogen atom

The initial energy \(E_1\) of the hydrogen atom in the \(n=3\) state is \( E_1= -13.6 / 3^2 eV = -1.51eV\)
02

Find the final energy of the system

The final energy \(E_2\) of the system is the sum of the final kinetic energy of the electron and the ionized hydrogen atom. For a hydrogen atom, the energy in the ionized state is 0eV. So we have \(E_2 = 0eV + 8.00eV = 8.00eV\)
03

Calculate the energy of the photon

The energy of the photon \(E_{\text{photon}}\) is the difference in initial and final energies of the electron (conservation of energy). So \(E_{\text{photon}} = E_2 - E_1 = 8.00eV - -1.51eV = 9.51eV\)
04

Convert energy to Joules

1 eV = \(1.602 \times 10^{-19} J\). So \(E_{\text{photon}} = 9.51eV \times 1.602 \times 10^{-19} J/eV = 1.52 \times 10^{-18} J\)
05

Relate energy to wavelength and solve for wavelength

Using the relation \(E = h \cdot c / \lambda\), where \(h\) is Planck's constant (\(6.626 \times 10^{-34}\) Js) and \(c\) is speed of light (\(2.998 \times 10^8\ ms^-1\)), we get \(\lambda = h \cdot c / E_{\text{photon}} = (6.626 \times 10^{-34} \times 2.998 \times 10^{8}) / 1.52 \times 10^{-18} = 1.30 \times 10^{-7} m\) or 130 nm.

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Key Concepts

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

Quantum Mechanics
Quantum Mechanics is a fundamental theory in physics that provides a description of the physical properties of nature at the scale of atoms and subatomic particles. It's a complex and often non-intuitive field that revolutionized our understanding of phenomena at small scales, leading to the development of technologies like transistors, lasers, and magnetic resonance imaging machines.

At the heart of quantum mechanics is the idea that energy and matter exhibit both particle-like and wave-like properties. In the context of the hydrogen atom absorption problem, quantum mechanics dictates that the electron can only inhabit certain discrete energy levels. When a photon with energy corresponding to the difference between these levels interacts with the atom, the electron can 'jump' to a higher level, or even escape the atom if the photon's energy is sufficient, a phenomenon known as ionization.

This description of discrete energy levels is a radical departure from the continuous energy variations described in classical physics and is essential to understanding the interaction between matter and light, as seen in this textbook problem.
Energy Levels of Hydrogen Atom
The Energy Levels of a Hydrogen Atom are quantized, meaning the electron bound to the hydrogen nucleus can only have certain specific values of energy. These specific energies are determined by the principal quantum number, denoted by the symbol 'n', and can be calculated using the formula \( E_n = -\frac{13.6 \mathrm{eV}}{n^2} \).

This equation arises from the Bohr model of the hydrogen atom, where \( E_n \) represents the energy of the electron in the 'n-th' energy level and 13.6 eV is the ionization energy of hydrogen. The negative sign indicates that these energy levels are below the energy of a free electron, which is defined as zero. According to quantum mechanics, transitions between these energy levels in an atom lead to absorption or emission of photons. The exercise showcases a situation where a photon is absorbed and the atom is ionized, bypassing all lower energy levels and releasing the electron into a free state.
Photoelectric Effect
The Photoelectric Effect is a phenomenon where electrons are ejected from a material when it is exposed to light of a certain minimum frequency. This effect was explained by Albert Einstein, who proposed that light consisted of packets of energy called photons. Each photon has an energy given by the equation \( E = h \cdot u \) where 'h' is Planck's constant and '\(u\)' is the frequency of the light. If the energy of the photon is equal to or greater than the work function (the minimum energy needed to remove an electron) of the material, then electrons will be emitted.

This concept is crucial in understanding how the hydrogen atom in the given textbook exercise absorbs a photon, leading to the ionization of the atom and release of an electron with kinetic energy. The photoelectric effect provides the conceptual bridge between the wave-like behavior of light and its particle-like interactions with matter, which underpin the principles of quantum mechanics and are exemplified in problems such as the ones students encounter in their textbooks.

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Most popular questions from this chapter

A beam of alpha particles is incident on a target of lead. A particular alpha particle comes in "head-on" to a particular lead nucleus and stops \(6.50 \times 10^{-14} \mathrm{~m}\) away from the center of the nucleus. (This point is well outside the nucleus.) Assume that the lead nucleus, which has 82 protons, remains at rest. The mass of the alpha particle is \(6.64 \times 10^{-27} \mathrm{~kg} .\) (a) Calculate the electrostatic potential energy at the instant that the alpha particle stops. Express your result in joules and in \(\mathrm{MeV}\). (b) What initial kinetic energy (in joules and in MeV) did the alpha particle have? (c) What was the initial speed of the alpha particle?

\(\mathrm{CP}\) (a) A particle with mass \(m\) has kinetic energy equal to three times its rest energy.What is the de Broglie wavelength of this particle? (Hint: You must use the relativistic expressions for momentum and kinetic energy: \(E^{2}=(p c)^{2}+\left(m c^{2}\right)^{2}\) and \(\left.K=E-m c^{2} .\right)\) (b) Determine the numerical value of the kinetic energy (in MeV) and the wavelength (in meters) if the particle in part (a) is (i) an electron and (ii) a proton.

Calculate the de Broglie wavelength of (a) a \(50.0 \mathrm{~kg}\) woman jogging leisurely at \(2.0 \mathrm{~m} / \mathrm{s},\) (b) a free electron with kinetic energy \(2.0 \mathrm{MeV}\). and (c) a free electron with kinetic energy \(20 \mathrm{eV}\). Use the proper relativistic expression when necessary.

How does the wavelength of a helium ion compare to that of an electron accelerated through the same potential difference? (a) The helium ion has a longer wavelength, because it has greater mass. (b) The helium ion has a shorter wavelength, because it has greater mass. (c) The wavelengths are the same, because the kinetic energy is the same. (d) The wavelengths are the same, because the electric charge is the same.

Suppose that the uncertainty of position of an electron is equal to the radius of the \(n=1\) Bohr orbit for hydrogen. Calculate the simultaneous minimum uncertainty of the corresponding momentum component, and compare this with the magnitude of the momentum of the electron in the \(n=1\) Bohr orbit. Discuss your results.

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