/*! 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 6 The photoelectric threshold wave... [FREE SOLUTION] | 91Ó°ÊÓ

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The photoelectric threshold wavelength of a tungsten surface is 272 \(\mathrm{nm}\) . Calculate the maximum kinetic energy of the electrons ejected from this tungsten surface by ultraviolet radiation of frequency \(1.45 \times 10^{15}\) Hz. Express the answer in electron volts.

Short Answer

Expert verified
The maximum kinetic energy is approximately 2.87 eV.

Step by step solution

01

Understand the Problem

We are given the threshold wavelength for tungsten's photoelectric effect and the frequency of the incident light. Our task is to find the maximum kinetic energy of the ejected electrons when light of the given frequency strikes the tungsten surface.
02

Calculate the Energy of Incident Photon

The energy of a photon is given by the formula \( E = h u \), where \( h = 6.63 \times 10^{-34} \text{ J s} \) is Planck's constant and \( u = 1.45 \times 10^{15} \text{ Hz} \) is the frequency of the radiation. Calculate \( E \): \[ E = (6.63 \times 10^{-34}) \times (1.45 \times 10^{15}) \]
03

Calculate the Work Function from Threshold Wavelength

The work function \( \phi \) is the minimum energy needed to remove an electron from the surface, computed using the threshold wavelength \( \lambda_0 = 272 \text{ nm} \). The work function is given by \( \phi = \frac{hc}{\lambda_0} \), where \( c = 3 \times 10^8 \text{ m/s} \). Convert \( \lambda_0 \) to meters and compute \( \phi \).
04

Apply the Photoelectric Equation

The photoelectric equation is \( KE_{max} = E - \phi \), where \( KE_{max} \) is the maximum kinetic energy of ejected electrons. Substitute the photon energy \( E \) and the work function \( \phi \) calculated in previous steps.
05

Convert Energy to Electron Volts

The kinetic energy calculated will be in joules. Convert it to electron volts using the conversion factor: \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \). Divide the kinetic energy in joules by this factor to obtain the energy in electron volts.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy Calculation
In the context of the photoelectric effect, we often need to calculate the kinetic energy of electrons that are ejected from a material when exposed to light. The key formula used for this calculation is the photoelectric equation: \[KE_{max} = E - \phi\]where \( KE_{max} \) is the maximum kinetic energy of the ejected electrons, \( E \) is the energy of the incoming photon, and \( \phi \) is the work function of the material. This equation tells us that the kinetic energy of the electrons depends on the difference between the incoming photon's energy and the energy needed to eject it.
• If the photon's energy is less than the work function, no electrons will be ejected.• If the photon's energy is greater, the remaining energy converts into kinetic energy.
All calculations should consider the precision of constants like Planck's constant to maintain accuracy.
Photon Energy
Photon energy, which is crucial in the photoelectric effect, is the energy carried by a photon and is determined using the formula:\[E = hu\]Here, \( h = 6.63 \times 10^{-34} \text{ J s} \) is Planck's constant and \( u \) (pronounced nu) represents the frequency of the associated radiation.

It's important to understand that photon energy is directly proportional to frequency. So, as the frequency of the radiation increases, the energy of each photon increases.
• For media that utilizes electromagnetic radiation like ultraviolet light, the frequency often determines the photon's capability to interact with materials.• Calculating the energy correctly is crucial for applications that rely on precise energy levels, such as spectroscopy and quantum computing.
Work Function
The work function \( \phi \) of a material signifies the minimum energy necessary to remove an electron from its surface. This concept is foundational in understanding which materials will readily emit electrons when exposed to particular wavelengths or frequencies of light.
The formula to compute the work function is:\[\phi = \frac{hc}{\lambda_0}\]where \( \lambda_0 \) is the threshold wavelength in meters, \( h \) is Planck's constant, and \( c \) is the speed of light.• A high work function means that more energy is needed to dislodge an electron, indicative of 'tighter' electron binding by the material.• Conversely, a low work function might suggest that electrons can be emitted with exposure to even lower-energy photons, a property exploited in technologies like solar panels.
Ultraviolet Radiation
Ultraviolet (UV) radiation is a type of electromagnetic radiation with a wavelength shorter than that of visible light, making it invisible to the human eye. Its frequency is higher than visible light, making UV photons more energetic and thus capable of causing photoelectric emission in many materials that might not respond to visible light.
• UV light is categorized into three types: UVA, UVB, and UVC, each with distinct energy levels and effects. • UVC, for example, has the most energy and can cause significant photoelectric emission. • Applications abound, from disinfecting surfaces to forensic analysis, due to its strong interaction with various substances and ability to excite electrons.
Electron Volts Conversion
Converting energy values between joules and electron volts (eV) is common in physics, especially in calculations involving atomic-scale phenomena.
An electron volt is defined as the amount of kinetic energy an electron gains when it accelerates through a potential difference of one volt.• The conversion factor is: \[1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}\]This conversion is essential when dealing with minute energies typical in quantum physics problems such as the photoelectric effect.
The use of electron volts helps simplify these expressively smaller quantities, as working in joules for such small energies can be unwieldy and cumbersome.• When converting from joules to electron volts, divide the joule value by the factor \( 1.602 \times 10^{-19} \).• Always ensure to maintain significant figures to ensure precision in scientific communication and calculations.

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Most popular questions from this chapter

Protons are accelerated from rest by a potential difference of 4.00 \(\mathrm{kV}\) and strike a metal target. If a proton produces one photon on impact, what is the minimum wavelength of the resulting x rays? How does your answer compare to the minimum wave-length if \(4.00-\mathrm{keV}\) electrons are used instead? Why do x-ray tubes use electrons rather than protons to produce x rays? use electrons rather than protons to produce x rays?

If a photon of wavelength 0.04250 nm strikes a free electron and is scattered at an angle of \(35.0^{\circ}\) from its original direction. find (a) the change in the wavelength of this photon; (b) the wave-length of the scattered light; (c) the change in energy of the photon (is it a loss or a gain?); (d) the energy gained by the electron.

A laser used to weld detached retinas emits light with a wavelength of 652 \(\mathrm{nm}\) in pulses that are 20.0 \(\mathrm{ms}\) in duration. The average power during each pulse is 0.600 \(\mathrm{W}\) . (a) How much energy is in each pulse in joules? In electron volts? (b) What is the energy of one photon in joules? In electron volts? (c) How many photons are in each pulse?

Consider Compton scattering of a photon by a moving electron. Before the collision the photon has wavelength \(\lambda\) and is moving in the \(+x\) -direction, and the electron is moving in the - \(x\) -direction with total energy \(E\) (including its rest energy \(m c^{2} )\) . The photon and electron collide head on. After the collision, both are moving in the \(-x\) -direction (that is, the photon has been scattered by \(180^{\circ}\) ). (a) Derive an expression for the wavelength \(\lambda^{\prime}\) of the scattered photon. Show that if \(E \gg m c^{2}\) , where \(m\) is the rest mass of the electron, your result reduces to $$ \Lambda^{\prime}=\frac{h c}{E}\left(1+\frac{m^{2} c^{4} \lambda}{4 h c E}\right) $$ (b) A beam of infrared radiation from a \(\mathrm{CO}_{2}\) laser \((\lambda=10.6 \mu \mathrm{m})\) collides head-on with a beam of electrons, each of total energy \(E=10.0 \mathrm{GeV}\left(1 \mathrm{GeV}=10^{9} \mathrm{eV}\right) .\) Calculate the wavelength \(\lambda^{\prime}\) of the scattered photons, assuming a \(180^{\circ}\) scattering angle. (c) What kind of scattered photons are these (infrared, microwave, ultraviolet, etc.)? Can you think of an application of this effect?

(a) Using the Bohr model, calculate the speed of the electron in a hydrogen atom in the \(n=1,2\) and 3 levels. (b) Calculate- the orbital period in each of these levels. (c) The average lifetime of the first excited level of a hydrogen atom is \(1.0 \times 10^{-8} \mathrm{s}\) . In the Bohr model, how many orbits does an electron in the \(n=2\) level complete before reurning to the ground level?

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