Ultraviolet light of wavelength 220 nm illuminates a tungsten surface and electrons are ejected. A stopping potential of \(1.1 \mathrm{V}\) is able to just prevent any of the ejected electrons from reaching the opposite electrode. What is the work function for tungsten?

Short Answer

Expert verified
Answer: The work function of tungsten is \(7.27 * 10^{-19}\) Joules.

Step by step solution

01

Identify and list given informations

We have the following informations: - Wavelength of ultraviolet light (λ): 220 nm - Stopping potential (V_stop): 1.1 V
02

Understand the photoelectric effect equation

The photoelectric effect equation is given by: \(E = W + K_{max}\) where: - E is the energy of the incident light - W is the work function - \(K_{max}\) is the maximum kinetic energy of the ejected electrons The goal is to find the work function (W) for tungsten.
03

Calculate the energy of the incident light

First, let's convert the wavelength of ultraviolet light λ from nanometers to meters: λ = 220 nm * 10^{-9} m/nm = 2.20 * 10^{-7} m Then, use the equation for energy (E) based on the speed of light (c), Planck's constant (h), and the wavelength (λ): \(E = \dfrac{hc}{λ}\) Substitute the values of h = 6.63 * 10^{-34} J*s and c = 3.00 * 10^8 m/s: \(E = \dfrac{(6.63 * 10^{-34} \text{J*s})(3.00 * 10^8 \text{m/s})}{(2.20 * 10^{-7} \text{m})}\) \(E = 9.03 * 10^{-19}\) J
04

Calculate the maximum kinetic energy of the ejected electrons

Using the stopping potential (V_stop) and the charge of an electron (e = 1.60 * 10^{-19} C), find the maximum kinetic energy (K_max): \(K_{max} = e * V_{stop}\) \(K_{max} = (1.60 * 10^{-19} \text{C}) * (1.1 \text{V})\) \(K_{max} = 1.76 * 10^{-19}\) J
05

Calculate the work function

Now, use the photoelectric effect equation to find the work function (W): \(E = W + K_{max}\) Isolate the work function (W): \(W = E - K_{max}\) Substitute the values of E and K_max found in Steps 3 and 4: \(W = (9.03 * 10^{-19} \text{J}) - (1.76 * 10^{-19} \text{J})\) \(W = 7.27 * 10^{-19}\) J The work function for tungsten is \(7.27 * 10^{-19}\) Joules.

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 \(180.0^{\circ} .\)
A surgeon is attempting to correct a detached retina by using a pulsed laser. (a) If the pulses last for \(20.0 \mathrm{ms}\) and if the output power of the laser is \(0.500 \mathrm{W},\) how much energy is in each pulse? (b) If the wavelength of the laser light is \(643 \mathrm{nm}\), how many photons are present in each pulse?
Follow the steps outlined in this problem to estimate the time lag (predicted classically but not observed experimentally) in the photoelectric effect. Let the intensity of the incident radiation be $0.01 \mathrm{W} / \mathrm{m}^{2} .\( (a) If the area of the atom is \)(0.1 \mathrm{nm})^{2},$ find the energy per second falling on the atom. (b) If the work function is 2.0 eV, how long would it take (classically) for enough energy to fall on this area to liberate one photoelectron? (c) Explain briefly, using the photon model, why this time lag is not observed.
A \(100-\) W lightbulb radiates visible light at a rate of about $10 \mathrm{W} ;$ the rest of the EM radiation is mostly infrared. Assume that the lightbulb radiates uniformly in all directions. Under ideal conditions, the eye can see the lightbulb if at least 20 visible photons per second enter a dark-adapted eye with a pupil diameter of \(7 \mathrm{mm}.\) (a) Estimate how far from the source the lightbulb can be seen under these rather extreme conditions. Assume an average wavelength of 600 nm. (b) Why do we not normally see lightbulbs at anywhere near this distance?
A particle collides with a hydrogen atom in the \(n=2\) state, transferring 15.0 eV of energy to the atom. As a result, the electron breaks away from the hydrogen nucleus. What is the kinetic energy of the electron when it is far from the nucleus?
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