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

A certain atom has an energy state 3.50 eV above the ground state. When excited to this state, the atom remains for 2.0 μs, on average, before it emits a photon and returns to the ground state. (a) What are the energy and wavelength of the photon? (b) What is the smallest possible uncertainty in energy of the photon?

Short Answer

Expert verified
(a) Energy = 3.50 eV, Wavelength = 355 nm. (b) Uncertainty in energy = 5.26×1029 J.

Step by step solution

01

Understanding the Energy of the Photon

Since the atom emits a photon to return to the ground state, the energy of the photon corresponds to the energy difference between these states. This is given as 3.50 eV.
02

Converting Energy Units

Convert the energy from electronvolts (eV) to joules (J) using the conversion factor: 1 eV=1.602×1019 J. Thus, 3.50 eV=3.50×1.602×1019 J=5.607×1019 J.
03

Calculating the Wavelength of the Photon

Use the formula relating energy and wavelength: E=hcλ, where h=6.626×1034 J·s is Planck's constant and c=3.00×108 m/s is the speed of light. Solving for λ, we have:λ=hcE=6.626×1034×3.00×1085.607×10193.55×107 m or 355 nm
04

Understanding Uncertainty in Energy

According to the energy-time uncertainty principle, ΔEΔt2, where =h2π. In this case, Δt=2.0×106 s.
05

Calculating the Uncertainty in Energy

Calculate ΔE using ΔEh4πΔt. Substituting the values gives:ΔE6.626×10344π×2.0×1065.26×1029 J

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.

Energy Levels
Energy levels in an atom represent the specific amounts of energy an electron can have. These levels define the possible states or orbits that electrons can occupy around a nucleus. Each energy level corresponds to a specific energy value, measured in electronvolts (eV) or joules (J).
When an electron moves between these levels, it must absorb or emit energy equal to the difference between these levels. In this exercise, the atom has an energy state that is 3.50 eV above the ground state. This means that for an electron to reach this excited state from the ground level, it needs to absorb energy equivalent to 3.50 eV.
Understanding energy levels is crucial in atomic physics because they dictate how atoms absorb and release energy, influencing chemical reactions and emission spectra.
Photon Emission
Photon emission occurs when an electron transitions from a higher energy level to a lower one, emitting energy as a photon in the process. The energy of this photon matches the energy difference between the initial and final states.
For the problem we're examining, the atom emits a photon as it transitions to the ground state from an excited state. This emitted photon carries 3.50 eV, the energy it has to release to return to the ground level.
To find the wavelength of an emitted photon, we can use the equation E=hcλ, where E is the photon energy, h is Planck's constant, and c is the speed of light. This relationship underscores the connection between a photon's energy and its wavelength, which is essential in understanding light and electromagnetic radiation.
Uncertainty Principle
The uncertainty principle is a fundamental concept in quantum mechanics, famously introduced by Werner Heisenberg. It states that certain pairs of physical properties, like position and momentum, cannot both be precisely measured simultaneously. In the context of this problem, it refers to energy and time.
The energy-time uncertainty principle is given by ΔEΔt2, where ΔE is the uncertainty in energy and Δt is the uncertainty in time. Here, is the reduced Planck's constant, equal to h2π.
For the excited atom, remaining on average for 2.0 μs before emitting a photon, Δt represents the average lifetime of the excited state. This principle is pivotal in quantum physics, indicating that there's a limit to the precision with which certain physical quantities can be known.
Electronvolt to Joules Conversion
In atomic physics, energy is often measured in electronvolts (eV). An electronvolt is a unit of energy equal to approximately 1.602×1019 joules. It is often used in the field because it is a convenient scale for processes involving electrons in atoms.
In the given exercise, the energy difference is presented as 3.50 eV. To convert it to joules for further calculations, we use the following conversion: 3.50 eV=3.50×1.602×1019 J, resulting in 5.607×1019 J.
This conversion is essential for expressing energy in the International System of Units (SI), making calculations consistent and comparable with other scientific measurements.

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

A scientist has devised a new method of isolating individual particles. He claims that this method enables him to detect simultaneously the position of a particle along an axis with a standard deviation of 0.12 nm and its momentum component along this axis with a standard deviation of 3.0 × 1025 kg m/s. Use the Heisenberg uncertainty principle to evaluate the validity of this claim.

A 10.0-g marble is gently placed on a horizontal tabletop that is 1.75 m wide. (a) What is the maximum uncertainty in the horizontal position of the marble? (b) According to the Heisenberg uncertainty principle, what is the minimum uncertainty in the horizontal velocity of the marble? (c) In light of your answer to part (b), what is the longest time the marble could remain on the table? Compare this time to the age of the universe, which is approximately 14 billion years. (Hint: Can you know that the horizontal velocity of the marble is exactly zero?)

(a) What is the smallest amount of energy in electron volts that must be given to a hydrogen atom initially in its ground level so that it can emit the Hα line in the Balmer series? (b) How many different possibilities of spectral-line emissions are there for this atom when the electron starts in the n = 3 level and eventually ends up in the ground level? Calculate the wavelength of the emitted photon in each case.

An electron beam and a photon beam pass through identical slits. On a distant screen, the first dark fringe occurs at the same angle for both of the beams. The electron speeds are much slower than that of light. (a) Express the energy of a photon in terms of the kinetic energy K of one of the electrons. (b) Which is greater, the energy of a photon or the kinetic energy of an electron?

A hydrogen atom is in a state with energy -1.51 eV. In the Bohr model, what is the angular momentum of the electron in the atom, with respect to an axis at the nucleus?

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