An isolated atom can emit a photon and the atom's internal energy drops. In fact, the process has a name:spontaneous emission. Can an isolated electron emit a photon? Why or why not?

Short Answer

Expert verified

An isolated electron cannot emit a photon on its own as this process would violate energy and momentum conservation.

Step by step solution

01

Concept used

An isolated electron cannot emit a photon on its own as this process would violate energy and momentum conservation. An electron has constant internal energy equal to mec2, where meis the rest mass of an electron and cis the velocity of light.

02

Photons and stationary electrons

If an electron emits a photon, its internal must drop increasing the kinetic energy of the photon. But, this cannot happen as the electron's internal energy is always a constant and cannot drop.

03

Conservation of momentum

Also, from the conservation of momentum, the final momentum of the electron and photon must be equal to the initial momentum of the electron. Considering the rest frame of the electron, the initial and final momentum of the electron is zero. This implies that the momentum of the photon after emission must be zero for momentum conservation. Again, this violates the known fact that photons always carry momentum and can never be zero.

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

A photon and an object of mass m have the same momentum p.

  1. Assuming that the massive object is moving slowly, so that non-relativistic formulas are valid, find in terms of m , p and c the ratio of the massive object’s kinetic energy, and argue that it is small.
  2. Find the ratio found in part (a), but using relativistically correct fomulas for the massive object. (Note: E2=p2c2+m2c4may be helpful.)
  3. Show that the low-speed limit of the ratio of part (b) agrees with part (a) and that the high-speed limit is 1.
  4. Show that at very high speed, the kinetic energy of a massive object approaches .

A stationary muon μ- annihilates with a stationary antimuonμ+ (same mass, role="math" localid="1657587173645" 1.88×10-28kg. but opposite charge). The two disappear, replaced by electromagnetic radiation. (a) Why is it not possible for a single photon to result? (b) Suppose two photons result. Describe their possible directions of motion and wavelengths.

A gamma-ray photon changes into a proton-antiproton pair. Ignoring momentum conservation, what must have been the wavelength of the photon (a) if the pair is stationary after creation, and (b) if each moves off at 0.6c, perpendicular to the motion of the photon? (c) Assume that these interactions occur as the photon encounters a lead plate and that a lead nucleus participates in momentum conservation. In each case, what fraction of the photon's energy must be absorbed by a lead nucleus?

You are conducting a photoelectric effect experiment by shining a light of 500 nmwavelength at a piece of metal and determining the stopping potential. If, unbeknownst to you, your 500 nm source actually contained a small amount of ultraviolet light, would it throw off your results by a small amount, or by quite a bit? Explain.

You are an early 20th-century experimental physicist and do not know the value of Planck's constant. By a suitable plot of the following data, and using Einstein's explanation of the photoelectric effect (KE=Wϕ. where his not known), determine Planck's constant.

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