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Light of wavelength \(200 \mathrm{nm}\) shines on an aluminum surface; \(4.20 \mathrm{eV}\) is required to eject an electron. What is the kinctic energy of (a) the fastest and (b) the slowest ejected electrons? (c) What is the stopping potential for this situation? (d) What is the cutoff wavelength for aluminum?

Short Answer

Expert verified
(a) 2.00 eV (b) 0 eV (c) 2.00 V (d) 295 nm.

Step by step solution

01

Calculate Energy of Incident Photons

The energy of the photons is determined using the photon energy formula \[ E = \frac{hc}{\lambda} \]where \( h = 6.626 \times 10^{-34} \text{ J s} \) is Planck's constant, \( c = 3 \times 10^8 \text{ m/s} \) is the speed of light, and \( \lambda = 200 \times 10^{-9} \text{ m} \) is the wavelength of the light.Substituting the given values, \[ E = \frac{(6.626 \times 10^{-34} \text{ J s})(3 \times 10^8 \text{ m/s})}{200 \times 10^{-9} \text{ m}} = 9.93 \times 10^{-19} \text{ J} \]Converting this energy to electron volts (1 eV = \(1.602 \times 10^{-19} J\)), \[ E = \frac{9.93 \times 10^{-19} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} = 6.20 \text{ eV} \].
02

Determine Kinetic Energy of Ejected Electrons

The maximum kinetic energy (K.E) of the electrons ejected is given by the photoelectric effect equation: \[ KE_{max} = \text{Energy of the incident photon} - \text{Work function} \]\[ KE_{max} = 6.20 \text{ eV} - 4.20 \text{ eV} = 2.00 \text{ eV} \]For the slowest ejected electrons, the kinetic energy would be theoretically zero if other mechanisms (like internal energy loss processes) are ignored.
03

Calculate the Stopping Potential

The stopping potential \( V_0 \) is the potential required to stop the fastest electrons, and is given by \[ KE = eV_0 \]where \( e = 1 \text{ eV} \).Thus, \[ V_0 = 2.00 \text{ V} \].
04

Find the Cutoff Wavelength

The cutoff wavelength \( \lambda_c \) is the maximum wavelength that can eject electrons and can be found by setting the kinetic energy to zero:\[ \text{Energy of the incident photon} = \text{Work function} \]\[ \frac{hc}{\lambda_c} = \text{Work function} \]Substituting the given values, \[ \lambda_c = \frac{(6.626 \times 10^{-34} \text{ J s})(3 \times 10^8 \text{ m/s})}{4.20 \times 1.602 \times 10^{-19} \text{ J}} \approx 295 \text{ nm} \].

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Key Concepts

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

Photon Energy
Photon energy is a central concept in understanding the photoelectric effect. It refers to the energy carried by a single photon, which is a quantum of electromagnetic radiation. To calculate the energy of a photon, we use the formula:
  • \[ E = \frac{hc}{\lambda} \]
  • Where \( h \) is Planck's constant \((6.626 \times 10^{-34} \text{ J s})\), \( c \) is the speed of light \((3 \times 10^8 \text{ m/s})\), and \( \lambda \) is the wavelength of the incident light.
In the given problem, the wavelength is \(200 \text{ nm}\) or \(200 \times 10^{-9} \text{ m}\). Substituting these values in, the energy of a photon hitting the aluminum surface was calculated to be approximately \(6.20 \text{ eV}\). This energy represents how much potential influence the photon has to cause an effect when it interacts with matter, like ejecting electrons.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. In the context of the photoelectric effect, when photons strike a metal surface, electrons are ejected, gaining kinetic energy in the process. The maximum kinetic energy of the ejected electrons can be calculated using:
  • \[ KE_{max} = E_{photon} - W \]
  • Where \( E_{photon} \) is the energy of the incident photon and \( W \) is the work function of the material (aluminum in this case).
The work function is the minimum energy required to eject an electron from the surface. In our example, the photon energy is \(6.20 \text{ eV}\) and the work function of aluminum is \(4.20 \text{ eV}\). Thus, the maximum kinetic energy of the electrons is \(2.00 \text{ eV}\). For the slowest ejected electrons, the kinetic energy can theoretically be zero if all the photon energy is used up in overcoming the work function.
Stopping Potential
Stopping potential is the voltage required to stop the most energetic ejected electrons from reaching the other side of the photoelectric cell. It directly measures the maximum kinetic energy of the electrons emitted. The relationship between stopping potential and kinetic energy is given by:
  • \[ KE = eV_0 \]
  • Where \( KE \) is kinetic energy and \( V_0 \) is the stopping potential, with \( e \) as the elementary charge.
In this problem, the maximum kinetic energy was found to be \(2.00 \text{ eV}\), translating to a stopping potential \( V_0 \) of \(2.00 \text{ V}\). This provides insight into the energies of electrons ejected by the photon and helps in understanding the efficiency of the photoelectric effect in various applications.
Cutoff Wavelength
The cutoff wavelength is a critical parameter in photoelectron emission. It refers to the longest wavelength of incident light that can effectively eject electrons from a metal surface. At this wavelength, the energy of the photons just barely overcomes the work function, leaving no extra energy to be converted into kinetic energy for the electrons.
  • \[ \frac{hc}{\lambda_c} = W \]
  • Where \( \lambda_c \) is the cutoff wavelength, \( W \) is the work function, and the formula equates to zero kinetic energy of the electrons.
For aluminum, using a work function of \(4.20 \text{ eV}\), the cutoff wavelength comes out to about \(295 \text{ nm}\). This means any photon with a wavelength longer than \(295 \text{ nm}\) won't have sufficient energy to eject electrons from the aluminum surface.

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Most popular questions from this chapter

(a) The smallest amount of energy needed to eject an electron from metallic sodium is \(2.28 \mathrm{eV}\). Does sodium show a photoelectric effect for red light, with \(\lambda=680 \mathrm{nm} ?\) (That is, does the light cause electron emission?) (b) What is the cutoff wavelength for photoelectric emission from sodium? (c) To what color does that wavelength correspond?

Under ideal conditions, a visual sensation can occur in the human visual system if light of wavelength \(550 \mathrm{nm}\) is absorbed by the eye's retina at a rate as low as 100 photons per second. What is the corresponding rate at which energy is absorbed by the retina?

You will find in Chapter 39 that electrons cannot move in definite orbits within atoms, like the planets in our solar system. To see why, let us try to "observe" such an orbiting electron by using a light microscope to measure the electron's presumed orbital position with a precision of, say, \(10 \mathrm{pm}\) (a typical atom has a radius of about \(100 \mathrm{pm}\) ). The wavelength of the light used in the microscope must then be about \(10 \mathrm{pm}\). (a) What would be the photon energy of this light? (b) How much energy would such a photon impart to an electron in a head-on collision? (c) What do these results tell you about the possibility of "viewing" an atomic electron at two or more points along its presumed orbital path? (Hint: The outer electrons of atoms are bound to the atom by energies of only a few electron-volts.)

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