7.1.5 The Photoelectric Effect

Question 1

The Photoelectric Effect

Answer 1

• Shining light on a metal ejects electrons, increasing the current registered on an ammeter.
  • Albert Einstein extended Planck’s quantum theory to explain the photoelectric effect.

Question 2

note

Answer 2

• In the photoelectric effect, shining light on a metal ejects electrons, increasing the current registered on an ammeter.
• Increasing the intensity of the light causes more electrons to be ejected, but does not change the kinetic energy of the ejected electrons.
• However, increasing the frequency of light causes the
  electrons to be ejected with higher kinetic energies.
• Below the threshold frequency (v0) from the metal, no electrons are ejected
• In 1905, Albert Einstein extended Planck’s quantum theory to explain the photoelectric effect.
• Einstein suggested that light was made up of discrete packets of energy, now called photons.
• The energy of a photon (E photon ) is equal to
  Planck’s constant (h) times the frequency of the photon ( v).
• In the photoelectric effect, an electron is only ejected if the photon has enough energy to overcome the attraction between the electron and the nucleus. If the photon has more energy than is required, the leftover energy is converted to kinetic energy.
• The energy required to eject an electron from a metal is equal to the energy of a photon of light at the threshold frequency (v0). This energy is the work function (ϕ) of the metal.
• Einstein’s explanation of the photoelectric effect meant that light could be thought of as being made up of particles which are capable of moving electrons. This helped to explain other light-driven reactions, such as vision.

Question 3

Which of the following statements correctly describes an observation of the photoelectric effect?

Answer 3

There is a frequency of light below which no electrons are emitted.

Question 4

A metal begins to emit electrons, as measured in an apparatus similar to the one Hertz used, when exposed to light at a wavelength of 342 nm (ultraviolet). What is the work function of this metal?

Answer 4

3.63 eV

Question 5

Which of the following metals (with their associated threshold wavelength) has the lowest work-function value?

Answer 5

Cs, 602 nm

Question 6

Potassium (K) has a work function value of 2.29 eV. What is the wavelength of light required to begin to generate a current in an apparatus like the one Hertz used with potassium metal?

Answer 6

543 nm

Question 7

Which of the following metals requires ultraviolet light to exhibit the photoelectric effect?

Answer 7

Ag, work function = 4.74 eV

Question 8

What is the most significant conclusion to be drawn from the observations of the photoelectric effect?

Answer 8

That light is contained in discrete packages of energy.

Question 9

Which of the following statements about Einstein’s explanation of the photoelectric effect is not correct?

Answer 9

Because light is packaged discretely, increasing the intensity of light will increase the number of photons hitting a metal and thus the likelihood that an electron will be ejected.

Question 10

Choose the most correct statement about the kinetic energy of electrons ejected by a metal exhibiting the photoelectric effect.

Answer 10

It is the energy of the photon minus the work function for the metal.

Question 11

What does the energy of the work-function value represent?

Answer 11

The amount of energy required to overcome the attractive forces between an electron and the bulk metal.

Question 12

The work-function value for uranium metal is 3.63 eV. This energy is achieved with photons at a frequency of 8.77 × 10^14 s−1 (near ultraviolet). If light of a lower frequency than this is applied, what will happen?

Answer 12

The current due to the photoelectric effect will disappear.