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

An orbiting satellite can charge by the photoelectric effect when sunlight ejects electrons from its outer surface. Satellites must be designed to minimize such charging because it can ruin the sensitive microelectronics. Suppose a satellite is coated with platinum, a metal with a very large function (ϕ=5.32eV). Find the longest wavelength of incident sunlight that can eject an electron from the platinum.

Short Answer

Expert verified

The value of longest wavelength of incident sunlight is 233 nm.

Step by step solution

01

Identification of the given data

The given data is listed below as-

The work function is ϕ=5.32eV

02

Significance of the photoelectric equation.

The photoelectric equation which is used to find the wavelength is given by-

hcλ=ϕ+Km

Here, the work function depends on the material and condition of the surface and not on the wavelength of incident light.

03

To determine a value of the longest wavelength of incident sunlight

The value of maximum kinetic energy of a photon is assumed as zero to find the longest wavelength of incident sunlight i.e., Kmax=0

Therefore, the photoelectric equation is given by-

hcλ=ϕ+Kmax

Now, Kmax=0

Therefore, hcλmax=ϕ

Substitute for, hc = 1240 eV.nm, ϕ=5.32nmin the above equation.

λmax=hcϕ=1240eV.nm5.32nm=233nm

Thus, the value of the longest wavelength of incident sunlight is 233 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

Question: You will find in Chapter 39 that electrons cannot move in definite orbits within atoms, like the planets in our solar system. To see why, let us try to “observe” such an orbiting electron by using a light microscope to measure the electron’s presumed orbital position with a precision of, say, 10pm(a typical atom has a radius of about localid="1663132292844" 100pm). The wavelength of the light used in the microscope must then be about 10pm. (a) What would be the photon energy of this light? (b) How much energy would such a photon impart to an electron in a head-on collision? (c) What do these results tell you about the possibility of “viewing” an atomic electron at two or more points along its presumed orbital path? (Hint:The outer electrons of atomsare bound to the atom by energies of only a few electron-volts.)

Figure 38-24 shows an electron moving through several regions where uniform electric potentials V have been set up. Rank the three regions according to the de Broglie wavelength of the electron there, greatest first.

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 meter was once defined as 1650763.73 wavelengths of the orange light emitted by a source containing krypton-86 atoms. What is the photon energy of that light?

At what rate does the Sun emit photons? For simplicity, assume that the Sun’s entire emission at the rate of 3.9×1026Wis at the single wavelength of 550 nm.

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