/*! 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 18 You are performing a photoelectr... [FREE SOLUTION] | 91Ó°ÊÓ

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You are performing a photoelectric effect experiment. Using a photocathode made of cesium, you first illuminate it with a green laser beam \((\lambda=514.5 \mathrm{nm})\) of power \(100 \mathrm{~mW}\). Next, you double the power of your laser beam, to \(200 \mathrm{~mW}\). How will the energies per electron of the electrons emitted by the cathode compare for the two cases?

Short Answer

Expert verified
Answer: Doubling the power of the laser beam does not change the energies per electron of the electrons emitted by the cesium photocathode. It only affects the number of emitted electrons.

Step by step solution

01

Find the energy of photons from the green laser beam emitting

To find the energy of the photons from the green laser beam, we can use the equation: \(E = \frac{hc}{\lambda}\) where \(E\) is the energy, \(h \approx 6.63 \times 10^{-34} \;\text{J} \cdot \text{s}\) is Planck's constant, \(c \approx 3.00 \times 10^8 \; \text{m} / \text{s}\) is the speed of light, and \(\lambda = 514.5 \times 10^{-9} \; \text{m}\) is the wavelength of the green laser beam. Plugging the values, we have: \(E = \frac{(6.63 \times 10^{-34} \;\text{J} \cdot \text{s})(3.00 \times 10^8 \; \text{m} / \text{s})}{514.5 \times 10^{-9} \; \text{m}}\) \(E \approx 3.87 \times 10^{-19} \; \text{J}\)
02

Calculate the number of photons emitted per second

We are given the power of the laser beam as 100 mW and 200 mW. We'll first calculate the number of photons emitted when the power is 100 mW. Power (P) = Energy (E) * Number of photons (n) / Time (t) n = P * t / E At 100 mW of power, we have: n = (100 \times 10^{-3} \; \text{J/s}) \times (\text{1 s}) / (3.87 \times 10^{-19} \; \text{J}) n \(\approx 2.58 \times 10^{17}\) photons/s
03

Calculate the number of emitted electrons in each case

As energy is conserved, one photon hits an electron in the photocathode, and thus we can assume the number of emitted electrons is equal to the number of photons for both the cases. Therefore, the number of emitted electrons when the power is 100 mW is: N1 = \(2.58 \times 10^{17}\) electrons/s Now, when the power is doubled to 200 mW: N2 = (200 \times 10^{-3} \; \text{J/s}) \times (\text{1 s}) / (3.87 \times 10^{-19} \; \text{J}) N2 \(\approx 5.16 \times 10^{17}\) electrons/s
04

Compare the energies per electron of the electrons emitted by the cathode

Since the energy of the photons is the same in both cases, the energy per emitted electron doesn't change as the individual energies of electrons remain same in both cases. Therefore, the energies per electron do not change when we double the power of the laser beam. Conclusion: Doubling the power of the laser beam does not change the energies per electron of the electrons emitted by the cathode; it only changes the number of emitted electrons.

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Key Concepts

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

Photon Energy
The energy of a photon is a crucial part of understanding the photoelectric effect, especially in experiments involving a laser beam illuminating a cesium photocathode. The formula used to find photon energy is \[ E = \frac{hc}{\lambda} \] where:
  • \( E \) is the photon energy,
  • \( h \approx 6.63 \times 10^{-34} \, \text{J} \cdot \text{s} \) is Planck's constant,
  • \( c \approx 3.00 \times 10^8 \, \text{m/s} \) is the speed of light,
  • \( \lambda \) is the wavelength of the laser light.
In this case, the wavelength of the green laser is given as 514.5 nm. Substituting this value into the formula allows us to calculate that each photon carries approximately \( 3.87 \times 10^{-19} \; \text{J} \) of energy. This energy influences the photoelectric effect by providing the necessary energy to release electrons from the metal surface when absorbed.
Laser Beam Power
Laser beam power is directly related to the intensity and amount of light energy the beam provides per unit of time. In the context of the photoelectric effect experiment:
  • Power is measured in watts (W), where 1 watt is equivalent to 1 joule per second.
With a laser power of 100 mW, the number of photons hitting the cesium surface per second can be calculated. Applying the formula:\[\text{Number of photons} = \frac{\text{Power} \times \text{Time}}{\text{Energy per photon}}\]we determine that approximately \( 2.58 \times 10^{17} \) photons strike the surface per second.Doubling the laser power to 200 mW simply doubles the number of photons, resulting in \( 5.16 \times 10^{17} \) photons per second. It's important to understand that while increasing the power increases the number of photons, each photon's energy remains constant in this scenario.
Electron Emission
In photoelectric experiments, electron emission occurs when a material, such as a cesium photocathode, is exposed to light photons. The energy from the photons is transferred to the electrons, providing enough energy to overcome the work function—the minimum energy needed to float an electron free. When photons of sufficient energy hit the photocathode, electrons are released.

In our experiment:
  • The number of electrons emitted corresponds to the number of photons striking the surface.
  • This means, more photons (due to increased power) lead to more emitted electrons.
However, the energy of individual electrons remains unchanged as it directly depends on the photon's energy, which remains constant even when the power of the laser beam is increased. Thus, increasing the beam power enhances the emission rate, not the energy per electron.
Cesium Photocathode
Cesium is often used as a photocathode in photoelectric effect experiments due to its relatively low work function. This property makes it easier for lower energy photons, such as green laser light, to induce electron emission. When photons from the laser beam strike the cesium photocathode:
  • Electrons are emitted if the photon's energy is greater than the work function of cesium.
  • This threshold ensures that not all sources of light can cause emission, highlighting the selective reactivity of cesium to light energy.
The cesium photocathode's efficient conversion of photon energy to electron emission makes it ideal for studying the photoelectric effect. By adjusting variables such as photon energy and laser power, a clearer understanding of light-material interactions can be gained.

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Most popular questions from this chapter

Suppose that Fuzzy, a quantum-mechanical duck, lives in a world in which Planck's constant \(\hbar=1.00 \mathrm{~J}\) s. Fuzzy has a mass of \(0.500 \mathrm{~kg}\) and initially is known to be within a \(0.750-\mathrm{m}-\) wide pond. What is the minimum uncertainty in Fuzzy's speed? Assuming that this uncertainty prevails for \(5.00 \mathrm{~s}\), how far away could Fuzzy be from the pond after 5.00 s?

A nocturnal bird's eye can detect monochromatic light of frequency \(5.8 \cdot 10^{14} \mathrm{~Hz}\) with a power as small as \(2.333 \cdot 10^{-17} \mathrm{~W}\). What is the corresponding number of photons per second a nocturnal bird's eye can detect?

An X-ray photon with an energy of \(50.0 \mathrm{keV}\) strikes an electron that is initially at rest inside a metal. The photon is scattered at an angle of \(45^{\circ} .\) What is the kinetic energy and momentum (magnitude and direction) of the electron after the collision? You may use the nonrelativistic relationship connecting the kinetic energy and momentum of the electron.

Scintillation detectors for gamma rays transfer the energy of a gamma-ray photon to an electron within a crystal, via the photoelectric effect or Compton scattering. The electron transfers its energy to atoms in the crystal, which re-emit it as a light flash detected by a photomultiplier tube. The charge pulse produced by the photomultiplier tube is proportional to the energy originally deposited in the crystal; this can be measured so an energy spectrum can be displayed. Gamma rays absorbed by the photoelectric effect are recorded as a photopeak in the spectrum, at the full energy of the gammas. The Compton-scattered electrons are also recorded, at a range of lower energies known as the Compton plateau. The highest-energy of these form the Compton edge of the plateau. Gamma-ray photons scattered \(180 .^{\circ}\) by the Compton effect appear as a backscatter peak in the spectrum. For gamma-ray photons of energy \(511 \mathrm{KeV}\) calculate the energies of the Compton edge and the backscatter peak in the spectrum.

In classical mechanics, for a particle with no net force on it, what information is needed in order to predict where the particle will be some later time? Why is this prediction not possible in quantum mechanics?

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