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

Question: Figure 38-13 shows a case in which the momentum component

pxof a particle is fixed so that px=0 ; then, from Heisenberg’s uncertainty principle (Eq. 38-28), the position x of the particle is completely unknown. From the same principle it follows that the opposite is also true; that is, if the position of a particle is exactly known (x=0), the uncertainty in its momentum is infinite.Consider an intermediate case, in which the position of aparticle is measured, not to infinite precision, but to within a distanceof λ2π, where λis the particle’s de Broglie wavelength.Show that the uncertainty in the (simultaneously measured) momentumcomponent is then equal to the component itself; that is,px=p. Under these circumstances, would a measured momentumof zero surprise you? What about a measured momentum of 0.5p? Of 0.2p? Of 12p?

Short Answer

Expert verified

It is proven that px=p

Yes, measure the momentum of zero, 0.5p,2pand 12pwould all be surprising.

Step by step solution

01

Concept used to solve the question.

Heisenberg’s Uncertainty Principle:

According to Heisenberg’s Uncertainty Principle, the uncertainty in position and momentum are related as,

xpxh2π

Where;

xis uncertainty in position along the axis.

pxis uncertainty in momentum along the axis.

his plank’s constant.

02

Finding The uncertainty in momentum 

From Heisenberg’s uncertainty Principle,

xpxh2π

For minimum uncertainty

xpx=h2π

Let, x=λ2π

λ2πpx=h2πpx=hλ

From de Broglie relation we know

p=hλ

Therefore,

px=p

Hence it is proved that px=p

Yes, measure momentum of zero, 0.5p,2pand 12pwould all be surprising.

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

What is the maximum wavelength shift for a Compton collision between a photon and a free photon?

Assuming that your surface temperature is98.6° Fand that you are an ideal blackbody radiator (you are close), find

(a) the wavelength at which your spectral radiancy is maximum,

(b) the power at which you emit thermal radiation in a wavelength range of 1.00nmat that wavelength, from a surface area of4.00 cm2, and

(c) the corresponding rate at which you emit photons from that area. Using a wavelength of500nm (in the visible range),

(d) recalculate the power and

(e) the rate of photon emission. (As you have noticed, you do not visibly glow in the dark.)

A photon undergoes Compton scattering off a stationary free electron.

The photon scatters at90.0° from its initial direction;

its initial wavelength is3×10-12m . What is the electron’s kinetic energy?

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

A light detector (your eye) has an area of 2.00×10-6m2and absorbs 80% of the incident light, which is at wavelength 500 nm. The detector faces an isotropic source, 3.00 m from the source. If the detector absorbs photons at the rate of exactly4.000s-1at what power does the emitter emit light?

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