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What new idea about light did Einstein use to explain the photoelectric effect? Why does the photoelectric effect exhibit a threshold frequency but not a time lag?

Short Answer

Expert verified
Einstein proposed the photon theory, stating that light is composed of particles called photons each with energy proportional to its frequency. The photoelectric effect exhibits a threshold frequency because photons must have enough energy to dislodge electrons, and there is no time lag because this energy transfer occurs instantaneously.

Step by step solution

01

Introduction

The photoelectric effect is the emission of electrons from a metal surface when light shines on it. Traditional wave theories of light failed to explain certain characteristics of this phenomenon, particularly the threshold frequency and the absence of a time lag.
02

Einstein's New Idea

Einstein proposed that light consists of particles called photons. Each photon has energy proportional to its frequency, given by \( E = h u \), where \( E \) is the energy, \( h \) is Planck's constant, and \( u \) is the frequency of the light.
03

Threshold Frequency

The threshold frequency is the minimum frequency of light required to emit electrons from a metal surface. According to Einstein's idea, for electrons to be emitted, the energy of a photon (determined by its frequency) must be at least equal to the work function of the metal. Therefore, if the frequency is below a certain threshold, photons don't have enough energy to dislodge electrons from the metal's surface.
04

Absence of Time Lag

In the photoelectric effect, electrons are emitted almost instantaneously when light above the threshold frequency shines on the metal. This lack of delay can be explained by photon theory: electrons absorb the energy of individual photons in an all-or-nothing process. If a photon has sufficient energy, it will immediately transfer energy to an electron, causing its ejection without any time lag.

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Key Concepts

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

Einstein's Photon Theory
Einstein revolutionized our understanding of light by proposing that it doesn't just behave as a wave, but also as a particle. These particles are called photons.
Each photon carries a specific amount of energy, which depends on its frequency. You can calculate this energy using the formula:
\[ E = h u \] where: \( E \) is the energy, \( h \) is Planck's constant, and \( u \) is the frequency of the light.
This means that light with a higher frequency (like ultraviolet light) will have higher-energy photons than light with a lower frequency (like visible light).
  • Photons interact with electrons in a metal surface to cause the emission
  • If a photon's energy is too low, it won't dislodge an electron
Understanding photons helps explain phenomena like the photoelectric effect, where light causes electrons to be ejected from a metal surface.
Threshold Frequency
The threshold frequency is crucial in the photoelectric effect. It's the minimum frequency of light needed to eject electrons from a metal surface.
Why is this important? Einstein's photon theory tells us that the energy of a photon is directly proportional to its frequency:
\[ E = h u \] For an electron to be emitted, the energy from a single photon must be at least equal to the work function of the metal. The work function is the minimum energy needed to dislodge an electron.

Here's what happens if the photon's frequency is below the threshold:
  • The photon's energy is too low to free an electron
  • No electrons are emitted, no matter how intense the light is
If the frequency is above the threshold, electrons are emitted almost instantly. This explains why only certain frequencies of light can trigger the photoelectric effect.
Planck's Constant
Planck's constant is a fundamental quantity in quantum mechanics. It's represented by the symbol \(h\) and has a value of approximately \[ 6.626 \times 10^{-34} \] joule-seconds.
It acts as a bridge between the energy of a photon and its frequency, as shown in the formula:
\[ E = h u \] Planck's constant tells us how much energy one photon carries based on its frequency. Here's why it's important in the photoelectric effect:
  • It helps calculate if a photon has enough energy to eject an electron
  • It ties the energy of light directly to its frequency, explaining the threshold frequency

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Most popular questions from this chapter

Compare the wavelengths of an electron (mass \(=9.11 \times 10^{-31} \mathrm{~kg}\) ) and a proton (mass \(=1.67 \times 10^{-27} \mathrm{~kg}\) ), each having (a) a speed of \(3.4 \times 10^{6} \mathrm{~m} / \mathrm{s} ;\) (b) a kinetic energy of \(2.7 \times 10^{-15} \mathrm{~J}\)

In order to comply with the requirement that energy be conserved, Einstein showed in the photoelectric effect that the energy of a photon \((h v)\) absorbed by a metal is the sum of the work function (\phi), which is the minimum energy needed to dislodge an electron from the metal's surface, and the kinetic energy \(\left(E_{\mathrm{k}}\right)\) of the electron: \(h \nu=\phi+E_{k}\). When light of wavelength \(358.1 \mathrm{nm}\) falls on the surface of potassium metal, the speed (u) of the dislodged electron is \(6.40 \times 10^{5} \mathrm{~m} / \mathrm{s}\) (a) What is \(E_{\mathrm{k}}\left(\frac{1}{2} \mathrm{mu}^{2}\right)\) of the dislodged electron? (b) What is \(\phi\) (in \(\mathrm{J}\) ) of potassium?

For each of the following sublevels, give the \(n\) and \(l\) values and the number of orbitals: (a) \(6 g ;(b) 4 s ;(c) 3 d\)

Covalent bonds in a molecule absorb radiation in the IR region and vibrate at characteristic frequencies. (a) The \(\mathrm{C}-\mathrm{O}\) bond absorbs radiation of wavelength \(9.6 \mu \mathrm{m}\). What frequency (in \(\mathrm{s}^{-1}\) ) corresponds to that wavelength? (b) The \(\mathrm{H}-\mathrm{Cl}\) bond has a frequency of vibration of \(8.652 \times 10^{13} \mathrm{~Hz}\). What wavelength (in \(\mu \mathrm{m}\) ) corresponds to that frequency?

Enormous numbers of microwave photons are needed to warm macroscopic samples of matter. A portion of soup containing \(252 \mathrm{~g}\) of water is heated in a microwave oven from \(20 .^{\circ} \mathrm{C}\) to \(98^{\circ} \mathrm{C}\), with radiation of wavelength \(1.55 \times 10^{-2} \mathrm{~m}\). How many photons are absorbed by the water in the soup?

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