(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 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

Nuclei in a radium-226 radioactive source emit photons whose energy is $186 \mathrm{keV} .$ These photons are scattered by the electrons in a metal target; a detector measures the energy of the scattered photons as a function of the angle \(\theta\) through which they are scattered. Find the energy of the \(\gamma\) -rays scattered through \(\theta=90.0^{\circ}\) and \(180.0^{\circ} .\)
Photons of energy \(E=4.000 \mathrm{keV}\) undergo Compton scattering. What is the largest possible change in photon energy, measured as a fraction of the incident photon's energy \(\left(E-E^{\prime}\right) / E ?\)
What potential difference must be applied to an x-ray tube to produce x-rays with a minimum wavelength of \(45.0 \mathrm{pm} ?\)
A photoelectric effect experiment is performed with tungsten. The work function for tungsten is 4.5 eV. (a) If ultraviolet light of wavelength \(0.20 \mu \mathrm{m}\) is incident on the tungsten, calculate the stopping potential. (b) If the stopping potential is turned off (i.e., the cathode and anode are at the same voltage), the \(0.20-\mu \mathrm{m}\) incident light produces a photocurrent of \(3.7 \mu \mathrm{A} .\) What is the photocurrent if the incident light has wavelength \(400 \mathrm{nm}\) and the same intensity as before?
(a) What is the difference in radius between the \(n=1\) state and the \(n=2\) state for hydrogen? (b) What is the difference in radius between the \(n=100\) state and the \(n=101\) state for hydrogen? How do the neighboring orbital separations compare for large and small \(n\) values?
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