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

Calculate the percentage change in photon energy during collision like that in Fig. 38-5 forϕ=90 and for radiation in

(a) the microwave range, withλ=3.0 cm ;

(b) the visible range, with λ=500 nm;

(c) the x-ray range, withλ=25 pm ; and

(d) the gamma-ray range, with a gamma photon energy of 1.0 MeV.

(e) What are your conclusions about the feasibility of detecting the Compton shift in these various regions of the electromagnetic spectrum, judging solely by the criterion of energy loss in a single photon-electron encounter?

Short Answer

Expert verified

(a) The percentage change in photon energy during collision in microwave range is8.1×109%.

(b) The percentage change in photon energy during collision in visible range is4.9×104%.

(c) The percentage change in photon energy during collision in ray range is8.9% .

(d) The percentage change in photon energy during collision in gamma ray range is66% .

(e) The Compton shift is zero for various of regions of electromagnetic spectrum.

Step by step solution

01

The percentage change in photon energy during collision and for radiation in microwave range.

(a)

The fractional change is written as follows:

ΔEE=Δhcλhcλ=λΔ1λ=λ1λ'1λ=λλ'1=λλ+Δλ1=1λΔλ+1=1λλC(1cosϕ)1+1

If

λ=3.0 cm=3.0×1010pm

andϕ=90

So,

ΔEE=13.0×10102.43(1cos90)1+1=8.1×1011=8.1×109%

Hence, the percentage change in photon energy during collision in microwave range is 8.1×109%.

02

The percentage change in photon energy during collision and for radiation in visible range. 

(b)

Let λ=500 nm=5.0×105pmand ϕ=90, thus it gives:

ΔEE=15.0×1052.43(1cos90)1+1=4.9×106=4.9×104%

Hence, the percentage change in photon energy during collision in visible range is 4.9×104%.

03

The percentage change in photon energy during collision and for radiation in x ray range. 

(c)

Letλ=25 pmand ϕ=90, thus it gives:

ΔEE=1252.43(1cos90)1+1=8.9×102=8.9%

Hence, the percentage change in photon energy during collision in x-ray range is8.9% .

04

The percentage change in photon energy during collision and for radiation in gamma ray range. 

(d)

Let

λ=hcE=1240 nm.eV1.0MeV=1.24×103nm=1.24pm

and , ϕ=90thus it gives:

ΔEE=11.242.43(1cos90)1+1=0.66=66%

Hence, the percentage change in photon energy during collision in gamma ray range is66%.

05

Conclusion about feasibility of detecting the Compton shift.

(e)

From the above calculation, the shorter the wavelength the greater the fractional energy change for the photon as a result of the Compton scattering. SinceΔEE is virtually zero for microwave and visible light, the Compton Effect is significant only in the x-ray to gamma ray range of the electromagnetic spectrum.

Hence, the Compton shift is zero for various of regions of electromagnetic spectrum.

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 metal plate is illuminated with light of a certain frequency. Which of the following determine whether or not electrons are ejected: (a) the intensity of the light, (b) how long the plate is exposed to the light, (c) the thermal conductivity of the plate, (d) the area of the plate, (e) the material of which the plate is made?

Question:For the arrangement of Figs. 38-14 and 38-15, electrons in the incident beam in region 1 have a speed of 1.60×107m/sand region 2 has an electric potential of V2-500V. What is the angular wave number in (a) region 1 and (b) region 2? (c) What is the reflection coefficient? (d) If the incident beam sends 3.00×109electrons against the potential step, approximately how many will be reflected?

Fig 38-14


Fig 38-15

Monochromatic light (that is, light of a single wavelength) is to be absorbed by a sheet of photographic film and thus recorded on the film. Photon absorption will occur if the photon energy equals or exceeds 0.6eV, the smallest amount of energy needed to dissociate an AgBr molecule in the film. (a) What is the greatest wavelength of light that can be recorded by the film? (b) In what region of the electromagnetic spectrum is this wavelength located?

A light detector has an absorbing area of 2.00×10-6m2 and absorbs 50% of the incident light, which is at a wavelength 600nm. The detector faces an isotropic source, 12.0 m from the source. The energy E emitted by the source versus time t is given in Fig. 38-26 ( Es=7.2nJ, ts=2.0s ). At what rate are photons absorbed by the detector?

The smallest amount of energy needed to eject an electron from metallic sodium is 2.28eV. Does sodium show a photoelectric effect for red light, with λ=680nm? (That is, does the light cause electron emission?) (b) What is the cutoff wavelength for photoelectric emission from sodium? (c) To what color does that wavelength correspond?

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