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Chapter 36: Problem 18

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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Most popular questions from this chapter

In a photoelectric effect experiment, a laser beam of unknown wavelength is shined on a cesium cathode (work function \(\phi=2.100 \mathrm{eV}\) ). It is found that a stopping potential of \(0.310 \mathrm{~V}\) is needed to eliminate the current. Next, the same laser is shined on a cathode made of an unknown material, and a stopping potential of \(0.110 \mathrm{~V}\) is found to be needed to eliminate the current. a) What is the work function for the unknown cathode? b) What would be a possible candidate for the material of this unknown cathode?

Two silver plates in vacuum are separated by \(1.0 \mathrm{~cm}\) and have a potential difference of \(20 . \mathrm{kV}\) between them. What is the largest wavelength of light that can be shined on the cathode to produce a current through the anode?

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?

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.

What is the minimum uncertainty in the velocity of a 1.0 -nanogram particle that is at rest on the head of a 1.0 -mm-wide pin?
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