/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 (a) Light of wavelength 300 nm i... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 27: Problem 12

(a) Light of wavelength 300 nm is incident on a metal that has a work function of 1.4 eV. What is the maximum speed of the emitted electrons? (b) Repeat part (a) for light of wavelength 800 nm incident on a metal that has a work function of \(1.6 \mathrm{eV} .\) (c) How would your answers to parts (a) and (b) vary if the light intensity were doubled?

Short Answer

Expert verified
Answer: The maximum speeds of the emitted electrons are 2.94 x 10^5 m/s for part (a) and 0 m/s for part (b). Doubling the light intensity does not change the maximum speeds of the electrons because it only increases the number of photons, not the energy of individual photons, which determines the speeds of the emitted electrons.

Step by step solution

01

Part (a)

1. Calculate the energy of the photons: \(E_{photon} = \dfrac{hc}{\lambda} = \dfrac{(6.626 \times 10^{-34} \ \text{Js})(3 \times 10^{8} \ \text{m/s})}{300 \times 10^{-9} \ \text{m}} = 6.626 \times 10^{-19} \ \text{J}\) 2. Compute the kinetic energy of the emitted electrons: \(K.E_{electron} = E_{photon} - W = 6.626 \times 10^{-19} \ \text{J} - 1.4 \ \text{eV} \times 1.6 \times 10^{-19} \ \text{J/eV} = 3.994 \times 10^{-19} \ \text{J}\) 3. Find the maximum speed of the emitted electrons: \(v = \sqrt{\dfrac{2K.E_{electron}}{m_{electron}}} = \sqrt{\dfrac{2(3.994 \times 10^{-19} \ \text{J})}{(9.109 \times 10^{-31} \ \text{kg})}} = 2.94 \times 10^5 \ \text{m/s}\) The maximum speed of the emitted electrons in part (a) is \(2.94 \times 10^5 \ \text{m/s}\).
02

Part (b)

1. Calculate the energy of the photons: \(E_{photon} = \dfrac{hc}{\lambda} = \dfrac{(6.626 \times 10^{-34} \ \text{Js})(3 \times 10^{8} \ \text{m/s})}{800 \times 10^{-9} \ \text{m}} = 2.481 \times 10^{-19} \ \text{J}\) 2. Compute the kinetic energy of the emitted electrons: \(K.E_{electron} = E_{photon} - W = 2.481 \times 10^{-19} \ \text{J} - 1.6 \ \text{eV} \times 1.6 \times 10^{-19} \ \text{J/eV} = 2.481 \times 10^{-19} \ \text{J} - 2.56 \times 10^{-19} \ \text{J}\) Since the photon energy is less than the work function of the metal, there will be no electrons emitted, and their speed will be zero. The maximum speed of the emitted electrons in part (b) is 0 m/s.
03

Part (c)

Doubling the light intensity means doubling the number of photons, but it does not change the energy of individual photons. The photoelectric effect depends on the energy of individual photons. Since the photons' energy remains the same in both cases, the electrons' maximum speed will also remain the same for parts (a) and (b). In conclusion, doubling the light intensity will not change the answers to parts (a) and (b).

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 91影视!

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 clean iron surface is illuminated by ultraviolet light. No photoelectrons are ejected until the wavelength of the incident UV light falls below 288 nm. (a) What is the work function (in electron-volts) of the metal? (b) What is the maximum kinetic energy for electrons ejected by incident light of wavelength \(140 \mathrm{nm} ?\)
A Compton scattering experiment is performed using an aluminum target. The incident photons have wavelength \(\lambda\). The scattered photons have wavelengths \(\lambda^{\prime}\) and energies \(E\) that depend on the scattering angle \(\theta.\) (a) At what angle \(\theta\) are scattered photons with the smallest energy detected? (b) At this same scattering angle \(\theta,\) what is the ratio \(\lambda^{\prime} / \lambda\) for \(\lambda=10.0 \mathrm{pm} ?\)
A hydrogen atom in its ground state is immersed in a continuous spectrum of ultraviolet light with wavelengths ranging from 96 nm to 110 nm. After absorbing a photon, the atom emits one or more photons to return to the ground state. (a) What wavelength(s) can be absorbed by the \(\mathrm{H}\) atom? (b) For each of the possibilities in (a), if the atom is at rest before absorbing the UV photon, what is its recoil speed after absorption (but before emitting any photons)? (c) For each of the possibilities in (a), how many different ways are there for the atom to return to the ground state?
An owl has good night vision because its eyes can detect a light intensity as faint as \(5.0 \times 10^{-13} \mathrm{W} / \mathrm{m}^{2} .\) What is the minimum number of photons per second that an owl eye can detect if its pupil has a diameter of \(8.5 \mathrm{mm}\) and the light has a wavelength of $510 \mathrm{nm} ?$
What is the ground state energy, according to Bohr theory, for (a) \(\mathrm{He}^{+},\) (b) \(\mathrm{Li}^{2+},\) (c) deuterium (an isotope of hydrogen whose nucleus contains a neutron as well as a proton)?
See all solutions

Recommended explanations on Physics Textbooks

Modern Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Energy Physics

Read Explanation

Particle Model of Matter

Read Explanation

Electric Charge Field and Potential

Read Explanation

Electrostatics

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.