Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 38: Q6P (page 1181)


What is the photon energy for yellow light from a highway sodium lamp at a wavelength of 589 nm?

Short Answer

Expert verified

The photon energy of the light is 2.1 eV.

Step by step solution

01

Describe the expression of energy of the photon

The energy of a photon of wavelength λis given by,

E=hcλ……. (1)

Here, h is Planck’s constant, and c is the speed of light.

02

Determine the photon energy for yellow light from a highway sodium lamp

Substitute the below values in eq 1.

λ=589nmh=6.626×10-34J.sc=3×108m/s

Therefore, the photon energy of the light is 2.1 eV.

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.

Start your free trial

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

Just after detonation, the fireball in a nuclear blast is approximately an ideal blackbody radiator with a surface temperature of about 1.0×107K.

(a) Find the wavelength at which the thermal radiation is maximum and (b) identify the type of electromagnetic wave corresponding to that wavelength. This radiation is almost immediately absorbed by the surrounding air molecules, which produces another ideal blackbody radiator with a surface temperature of about 1.0×105K.

(c) Find the wavelength at which the thermal radiation is maximum and (d) identify the type of electromagnetic wave corresponding to that wavelength.

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.)

Light strikes a sodium surface, causing photoelectric emission. The stopping potential for the ejected electrons is 50 V, and the work function of sodium is 2.2 eV. What is the wavelength of the incident light?

Question: In Eq. keep both terms, putting A=B=ψ. The

equation then describes the superposition of two matter waves of

equal amplitude, traveling in opposite directions. (Recall that this

is the condition for a standing wave.) (a) Show that |♆x,t|2 is

then given by |♆(x,t)|2=2ψ02[1+cos2kx]

(b) Plot this function, and demonstrate that it describes the square

of the amplitude of a standing matter wave. (c) Show that thenodes of this standing wave are located at x=(2n+1)(14λ),where n=0,1,2,3,…

and λ is the de Broglie wavelength of the particle. (d) Write a similar

expression for the most probable locations of the particle.

The highest achievable resolving power of a microscope is limited only by the wavelength used; that is, the smallest item that can be distinguished has dimensions about equal to the wavelength. Suppose one wishes to “see” inside an atom. Assuming the atom to have a diameter of 100pm, this means that one must be able to resolve a width of, say, 10pm.

(a) If an electron microscope is used, what minimum photon energy is required?

(b) If a light microscope is used, what minimum photon energy is required?

(c) Which microscope seems more practical? Why?
See all solutions

Recommended explanations on Physics Textbooks

Scientific Method Physics

Read Explanation

Wave Optics

Read Explanation

Nuclear Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation
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