Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 28: Problem 51

Many lasers, including the helium-neon, can produce beams at more than one wavelength. Photons can stimulate emission and cause transitions between the \(20.66-\mathrm{eV}\) metastable state and several different states of lower energy. One such state is 18.38 eV above the ground state. What is the wavelength for this transition? If only these photons leave the laser to form the beam, what color is the beam?

Short Answer

Expert verified
Answer: The wavelength of the emitted photon is approximately 542.8 nm, and the color of the beam is green.

Step by step solution

01

Calculate the energy difference between metastable and lower energy states

The energy difference between the metastable state (E1) and the lower energy state (E2) can be determined by the subtraction E1 - E2. Given E1 = 20.66 eV and E2 = 18.38 eV, the energy difference is: ΔE = E1 - E2 = 20.66 eV - 18.38 eV = 2.28 eV
02

Convert energy difference to Joules

In order to use Planck's equation to determine the wavelength of the emitted photon, we need to convert the energy difference from electron volts (eV) to Joules (J). We can use the following conversion factor: 1 eV = 1.602 x 10^{-19} J So, ΔE in Joules is: ΔE (J) = 2.28 eV × (1.602 × 10^{-19} J/eV) ≈ 3.6536 × 10^{-19} J
03

Calculate the wavelength using Planck's equation

Planck's equation relates the energy of a photon (E) to its wavelength (λ), through the equation: E = hc/λ Here, h is Planck's constant (6.626 x 10^{-34} Js), and c is the speed of light (3.00 x 10^8 m/s). To find the wavelength, we can rearrange the equation and plug in the energy value: λ = hc/ΔE λ ≈ (6.626 × 10^{-34} Js × 3.00 × 10^8 m/s) / (3.6536 × 10^{-19} J) ≈ 5.428 × 10^{-7} m
04

Convert wavelength to nanometers

It is more common to express wavelengths in nanometers (nm) rather than meters. To convert from meters to nanometers, multiply by 1 × 10^9: λ (nm) = 5.428 × 10^{-7} m × (1 × 10^9 nm/m) ≈ 542.8 nm
05

Determine the color of the beam

Now that we have the wavelength of the emitted photon, we can use the visible light spectrum to determine the color of the beam. Wavelengths of visible light range from approximately 400 nm (violet) to 700 nm (red). The given wavelength of 542.8 nm falls within the green portion of the visible spectrum. Hence, the color of the beam emitted by the helium-neon laser is green.

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

If diffraction were the only limitation on resolution, what would be the smallest structure that could be resolved in an electron microscope using 10-keV electrons?
A light-emitting diode (LED) has the property that electrons can be excited into the conduction band by the electrical energy from a battery; a photon is emitted when the electron drops back to the valence band. (a) If the band gap for this diode is \(2.36 \mathrm{eV},\) what is the wavelength of the light emitted by the LED? (b) What color is the light emitted? (c) What is the minimum battery voltage required in the electrical circuit containing the diode?
In the Davisson-Germer experiment (Section \(28.2),\) the electrons were accelerated through a \(54.0-\mathrm{V}\) potential difference before striking the target. (a) Find the de Broglie wavelength of the electrons. (b) Bragg plane spacings for nickel were known at the time; they had been determined through x-ray diffraction studies. The largest plane spacing (which gives the largest intensity diffraction maxima) in nickel is \(0.091 \mathrm{nm} .\) Using Bragg's law [Eq. ( \(25-15\) )], find the Bragg angle for the first-order maximum using the de Broglie wavelength of the electrons. (c) Does this agree with the observed maximum at a scattering angle of \(130^{\circ} ?\) [Hint: The scattering angle and the Bragg angle are not the same. Make a sketch to show the relationship between the two angles.]
An electron is confined to a box of length \(1.0 \mathrm{nm} .\) What is the magnitude of its momentum in the \(n=4\) state?
What is the electronic configuration of the ground state of the carbon atom? Write it in the following ways: (a) using spectroscopic notation \(\left(1 s^{2}, \ldots\right) ;\) (b) listing the four quantum numbers for each of the electrons. Note that there may be more than one possibility in (b).
See all solutions

Recommended explanations on Physics Textbooks

Translational Dynamics

Read Explanation

Turning Points in Physics

Read Explanation

Rotational Dynamics

Read Explanation

Modern Physics

Read Explanation

Electrostatics

Read Explanation

Medical Physics

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