A helium-neon laser is pumped by electric discharge. What wavelength electromagnetic radiation would be needed to pump it? See Figure 30.39 for energy-level information.

Short Answer

Expert verified

The needed electromagnetic radiation wavelength is 60.18 nm.

Step by step solution

01

Determine the formulas:

Consider the formula for the energy of X ray photons as follows:

\({\bf{E = }}\frac{{{\bf{hc}}}}{{\bf{\lambda }}}\)

Here,

λ = Wavelength

E = Energy of the x-ray photons

h = Planck's constant

c = Speed of light

02

Determine wavelength of electromagnetic radiation

Consider the given data:

\(\begin{array}{l}h = 6.626 \times {10^{ - 34}}\;\;{\rm{J}} \cdot s\\E = 20.66\;{\rm{eV}}\\c = 3 \times {10^8}\;\frac{{\rm{m}}}{{\rm{s}}}\end{array}\)

Substituting the values in equation of energy.

\(\begin{array}{l}20.66\;{\rm{eV}} = \frac{{\left( {6.626 \times {{10}^{ - 34}}\;{\rm{J}} \cdot {\rm{s}}} \right)\left( {3 \times {{10}^8}\;\frac{{\rm{m}}}{{\rm{s}}}} \right)}}{\lambda }\\\lambda = \frac{{\left( {6.626 \times {{10}^{ - 34}}\;{\rm{J \times s}}} \right)\left( {3 \times {{10}^8}\;\frac{{\rm{m}}}{{\rm{s}}}} \right)}}{{20.66\;{\rm{eV}}}}\\\lambda = 60.18 \times {10^{ - 9}}\;{\rm{m}}\\\lambda = 60.18\;{\rm{nm}}\end{array}\)

Consider the wavelength of the electromagnetic radiation is 60.18 nm.

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!

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

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