Essential Physics

Section 3 review

Is light a wave or a particle? This question confounded physicists for many centuries. In the 19th century, physicists were fairly certain that light was a wave, because they knew from Young’s double slit experiment that light can diffract and then interfere with itself. But the photoelectric effect, discovered at the end of the century, could not be explained by the wave model. Einstein soon resolved the problem of the photoelectric effect by explaining that light acts as little particles called photons, based on Planck’s equation for their energy, E = hf. While light can exhibit wave properties, at the atomic level light often exhibits properties of a particle. Quantum physics is the study of phenomena at the atomic and nuclear level where many physical properties are quantized, or only come in discrete values. An early success of the quantum theory was in explaining the photoelectric effect through quantized energy of photons. Read the text aloud

Read the text aloud

polarization, diffraction pattern, spectrograph, photon, Planck’s constant, quantum physics, photoelectric effect, threshold frequency, work function, detector, pixel, optical fibers

E = h f

E_{k} = h f - W_{0}

Review problems and questions

Describe evidence that light behaves like a particle and evidence that light behaves like a wave.

1. What is the energy of a photon of light that has a wavelength of λ = 2 μm?
2. What is the energy of a photon of light that has λ = 0.5 μm?
3. What part of the electromagnetic spectrum does each photon correspond to?

Answer:

9.9×10⁻²⁰ J
3.96×10⁻¹⁹ J
The light in a corresponds to the infrared part of the electromagnetic spectrum. The light in b corresponds to the visible part of the electromagnetic spectrum.

Solution:

The frequency for a photon of light that has a wavelength of 2 μm or 2×10⁻⁶ m is $\begin{matrix} c = f λ & \Rightarrow & \frac{1}{λ} c = f λ \frac{1}{λ} & \Rightarrow & f = \frac{c}{λ} = \frac{3 \times 10^{8} m/s}{2 \times 10^{- 6} m} = 1.5 \times 10^{14} Hz \end{matrix}$ Then use Planck's equation to find the energy of that photon: $E = h f = (6.63 \times 10^{- 34} J s) (1.5 \times 10^{14} Hz) = 9.9 \times 10^{- 20} J$
This light has one-fourth the wavelength, so it will have four times higher energy, or 3.96×10⁻¹⁹ J.
The light in a corresponds to the infrared part of the electromagnetic spectrum. The light in b corresponds to the visible part of the electromagnetic spectrum.

Explain how to find the maximum kinetic energy of an electron ejected from a material via the photoelectric effect.

A researcher was designing an experiment to demonstrate the photoelectric effect using the metal nickel, which has a work function of 5.15 eV. The electron volt, or eV, is a more convenient way to express photon energies than using joules. 1 eV = 1.602×10⁻¹⁹ J.
1. Explain what is meant physically by the work function.
2. Calculate the threshold frequency for nickel. To what region of the electromagnetic spectrum does this correspond?
3. Will light at an energy of 7.35 eV result in the production of photoelectrons from nickel? In what region of the electromagnetic spectrum is this light?
4. Will infrared light at a wavelength of 1 μm produce photoelectrons?
5. Will violet light (400 nm) produce photoelectrons?
6. Will ultraviolet light at 200 nm produce photoelectrons?

Answer:

The work function corresponds to the minimum energy required to liberate an electron from the metal.
The threshold frequency is 1.24×10¹⁵ Hz, which is ultraviolet light.
Yes; photoelectrons will be produced. This is ultraviolet light.
No.
No.
Yes.

Solution:

The work function corresponds to the minimum energy required to liberate an electron from the metal.
The energy of the work function is $W_{0} = (5.15 eV) (\frac{1.602 \times 10^{- 19} J}{1 eV}) = 8.25 \times 10^{- 19} J$ Electrons will begin to be released from the metal when the energy of the incident light is greater than the work function. The energy of the light is hf, so equate this to the work function W₀ and solve for the threshold frequency: $\begin{matrix} W_{0} = h f & \Rightarrow & f = \frac{W_{0}}{h} \end{matrix} = \frac{8.25 \times 10^{- 19} J}{6.63 \times 10^{- 34} J s} = 1.24 \times 10^{15} Hz$ Comparing this to the electromagnetic spectrum charts, we find that this is ultraviolet light.
An energy of 7.35 eV is greater than the work function of 5.15 eV, so photoelectrons will be produced. This energy is within the ultraviolet region of the spectrum since its frequency is $f = (\frac{7.35 eV}{5.15 eV}) (1.24 \times 10^{15} Hz) = 1.77 \times 10^{15} Hz$
We know from b that the threshold frequency is in the ultraviolet. Infrared light has a lower frequency and energy than ultraviolet light, so it will not produce photoelectrons.
Visible light, including violet light, has a lower frequency and energy than ultraviolet light, so it will not produce photoelectrons.
To answer this question, we must determine the frequency corresponding to ultraviolet light of wavelength λ = 200 nm: $f = \frac{c}{λ} = \frac{3 \times 10^{8} m/s}{200 \times 10^{- 9} m} = 1.5 \times 10^{15} Hz$ This frequency is higher than the threshold frequency, so it will produce photoelectrons.

Many forms of modern communication use wave pulses of infrared and visible light to transfer digital information. Describe why infrared (IR) or visible light, rather than radio frequencies, is used for transmitting the signals needed for high-definition video.


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