Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 32

How is the energy conserved when an atom makes a transition from a higher to a lower energy state?

Short Answer

Expert verified
When an atom makes a transition from a higher to a lower energy state, the energy conservation is maintained through the emission of a photon. The energy lost by the electron during the transition, denoted as ΔE, is equal to the difference between the higher (E1) and lower (E2) energy levels. The energy of the emitted photon (E_photon) is also equal to ΔE. This is demonstrated by the equation: \(E_1 - E_2 = h\nu\), where \(h\) is Planck's constant and \(\nu\) is the frequency of the photon. Thus, the energy conservation is achieved by balancing the energy lost by the electron with the energy gained by the emitted photon.

Step by step solution

01

Understanding energy levels in atoms

In atoms, electrons can only occupy specific energy levels. When an electron transitions from a higher energy level (E1) to a lower one (E2), it loses an amount of energy equal to the difference between the two levels, which is often denoted as ΔE.
02

Conservation of energy in atomic transitions

When an electron transitions from a higher to a lower energy level, the lost energy must be conserved. This conservation is realized by the emission of a photon, which is an energy-carrying particle. The energy of the emitted photon (E_photon) is equal to the difference in energy levels (ΔE) of the electron.
03

Calculating the energy of an emitted photon

Using the Planck-Einstein relation, we can relate the energy of the emitted photon to its frequency. The equation is : \[E_{photon} = h\nu\] where \(E_{photon}\) is the energy of the photon, \(h\) is Planck's constant (\(6.626 \times 10^{-34} Js\)), and \(\nu\) is the frequency of the photon.
04

Conservation of energy during the transition

To conserve energy during the transition between energy levels, the energy lost by the electron should be equal to the energy gained by the emitted photon. Therefore, we have: \[\Delta E = E_{photon}\] or \[E_1 - E_2 = h\nu\] This equation demonstrates that when an electron transitions from a higher energy level (E1) to a lower energy level (E2), the lost energy (ΔE) is conserved by the emission of a photon with energy equal to the difference between the two energy levels.

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

(a) Calculate the wavelength of a photon that has the same momentum as a proton moving with \(1 \%\) of the speed of light in a vacuum. (b) What is the energy of this photon in MeV? (c) What is the kinetic energy of the proton in MeV?

At what velocity will an electron have a wavelength of \(1.00 \mathrm{m} ?\)

Show that Stefan's law results from Planck's radiation law. Hint: To compute the total power of black body radiation emitted across the entire spectrum of wavelengths at a given temperature, integrate Planck's law over the entire spectrum \(P(T)=\int_{0}^{\infty} I(\lambda, T) d \lambda .\) Use the substitution \(x=h c / \lambda k T\) and the tabulated value of the integral \(\int_{0}^{\infty} d x x^{3} /\left(e^{x}-1\right)=\pi^{4} / 15\).

Calculate the velocity of a 1.0 - \(\mu\) m electron and a potential difference used to accelerate it from rest to this velocity.

The work function for barium is 2.48 eV. Find the maximum kinetic energy of the ejected photo electrons when the barium surface is illuminated with: (a) radiation emitted by a 100 -kW radio station broadcasting at 800 \(\mathrm{kHz} ;\) (b) a 633 -nm laser light emitted from a powerful HeNe laser; and (c) a 434-nm blue light emitted by a small hydrogen gas discharge tube.
See all solutions

Recommended explanations on Physics Textbooks

Waves Physics

Read Explanation

Fields in Physics

Read Explanation

Astrophysics

Read Explanation

Fluids

Read Explanation

Rotational Dynamics

Read Explanation

Magnetism and Electromagnetic Induction

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