A metal foil has a threshold frequency of 5.45 x 10^14 Hz.

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A metal foil has a threshold frequency of 5.45 x 10^14 Hz. Which of the colors of visible light have enough energy to eject electrons from this metal? – red – blue – indigo – green – violet – orange – yellow Question Source: McQuarrie, Rocki, and Gallogly’s General Chemistry

The Correct Answer and Explanation is:

To determine which colors of visible light can eject electrons from a metal foil with a threshold frequency of 5.45 x 10^14 Hz, we need to compare the energy of the photons corresponding to the different colors of visible light to the energy required to eject electrons from the metal.

The energy of a photon can be calculated using the formula:E=h⋅fE = h \cdot fE=h⋅f

where:

  • EEE is the energy of the photon (in joules),
  • hhh is Planck’s constant (6.626×10−346.626 \times 10^{-34}6.626×10−34 J·s),
  • fff is the frequency of the light (in Hz).

For the metal to eject an electron, the energy of the incoming photon must be greater than or equal to the work function of the metal, which corresponds to the threshold frequency.

Step 1: Calculate the work function energy of the metal

We are given that the threshold frequency fthreshold=5.45×1014 Hzf_{\text{threshold}} = 5.45 \times 10^{14} \, \text{Hz}fthreshold​=5.45×1014Hz. Using Planck’s constant:Ethreshold=h⋅fthreshold=(6.626×10−34 J\cdotps)×(5.45×1014 Hz)=3.61×10−19 JE_{\text{threshold}} = h \cdot f_{\text{threshold}} = (6.626 \times 10^{-34} \, \text{J·s}) \times (5.45 \times 10^{14} \, \text{Hz}) = 3.61 \times 10^{-19} \, \text{J}Ethreshold​=h⋅fthreshold​=(6.626×10−34J\cdotps)×(5.45×1014Hz)=3.61×10−19J

So, the work function energy is 3.61×10−19 J3.61 \times 10^{-19} \, \text{J}3.61×10−19J.

Step 2: Determine the energy of photons for different colors

Next, we will calculate the energy of photons for each color of visible light. The energy depends on the frequency of the light, so we need to know the frequency of each color. Below are approximate frequencies for the colors of visible light:

  • Red: 4.30×1014 Hz4.30 \times 10^{14} \, \text{Hz}4.30×1014Hz
  • Orange: 4.85×1014 Hz4.85 \times 10^{14} \, \text{Hz}4.85×1014Hz
  • Yellow: 5.10×1014 Hz5.10 \times 10^{14} \, \text{Hz}5.10×1014Hz
  • Green: 5.50×1014 Hz5.50 \times 10^{14} \, \text{Hz}5.50×1014Hz
  • Blue: 6.00×1014 Hz6.00 \times 10^{14} \, \text{Hz}6.00×1014Hz
  • Indigo: 6.20×1014 Hz6.20 \times 10^{14} \, \text{Hz}6.20×1014Hz
  • Violet: 7.50×1014 Hz7.50 \times 10^{14} \, \text{Hz}7.50×1014Hz

Now, calculate the energy of photons for each color:

  1. Red:

Ered=h⋅fred=(6.626×10−34 J\cdotps)×(4.30×1014 Hz)=2.85×10−19 JE_{\text{red}} = h \cdot f_{\text{red}} = (6.626 \times 10^{-34} \, \text{J·s}) \times (4.30 \times 10^{14} \, \text{Hz}) = 2.85 \times 10^{-19} \, \text{J}Ered​=h⋅fred​=(6.626×10−34J\cdotps)×(4.30×1014Hz)=2.85×10−19J

  1. Orange:

Eorange=h⋅forange=(6.626×10−34 J\cdotps)×(4.85×1014 Hz)=3.21×10−19 JE_{\text{orange}} = h \cdot f_{\text{orange}} = (6.626 \times 10^{-34} \, \text{J·s}) \times (4.85 \times 10^{14} \, \text{Hz}) = 3.21 \times 10^{-19} \, \text{J}Eorange​=h⋅forange​=(6.626×10−34J\cdotps)×(4.85×1014Hz)=3.21×10−19J

  1. Yellow:

Eyellow=h⋅fyellow=(6.626×10−34 J\cdotps)×(5.10×1014 Hz)=3.38×10−19 JE_{\text{yellow}} = h \cdot f_{\text{yellow}} = (6.626 \times 10^{-34} \, \text{J·s}) \times (5.10 \times 10^{14} \, \text{Hz}) = 3.38 \times 10^{-19} \, \text{J}Eyellow​=h⋅fyellow​=(6.626×10−34J\cdotps)×(5.10×1014Hz)=3.38×10−19J

  1. Green:

Egreen=h⋅fgreen=(6.626×10−34 J\cdotps)×(5.50×1014 Hz)=3.64×10−19 JE_{\text{green}} = h \cdot f_{\text{green}} = (6.626 \times 10^{-34} \, \text{J·s}) \times (5.50 \times 10^{14} \, \text{Hz}) = 3.64 \times 10^{-19} \, \text{J}Egreen​=h⋅fgreen​=(6.626×10−34J\cdotps)×(5.50×1014Hz)=3.64×10−19J

  1. Blue:

Eblue=h⋅fblue=(6.626×10−34 J\cdotps)×(6.00×1014 Hz)=3.98×10−19 JE_{\text{blue}} = h \cdot f_{\text{blue}} = (6.626 \times 10^{-34} \, \text{J·s}) \times (6.00 \times 10^{14} \, \text{Hz}) = 3.98 \times 10^{-19} \, \text{J}Eblue​=h⋅fblue​=(6.626×10−34J\cdotps)×(6.00×1014Hz)=3.98×10−19J

  1. Indigo:

Eindigo=h⋅findigo=(6.626×10−34 J\cdotps)×(6.20×1014 Hz)=4.11×10−19 JE_{\text{indigo}} = h \cdot f_{\text{indigo}} = (6.626 \times 10^{-34} \, \text{J·s}) \times (6.20 \times 10^{14} \, \text{Hz}) = 4.11 \times 10^{-19} \, \text{J}Eindigo​=h⋅findigo​=(6.626×10−34J\cdotps)×(6.20×1014Hz)=4.11×10−19J

  1. Violet:

Eviolet=h⋅fviolet=(6.626×10−34 J\cdotps)×(7.50×1014 Hz)=4.97×10−19 JE_{\text{violet}} = h \cdot f_{\text{violet}} = (6.626 \times 10^{-34} \, \text{J·s}) \times (7.50 \times 10^{14} \, \text{Hz}) = 4.97 \times 10^{-19} \, \text{J}Eviolet​=h⋅fviolet​=(6.626×10−34J\cdotps)×(7.50×1014Hz)=4.97×10−19J

Step 3: Compare the photon energies to the work function

To eject electrons, the energy of the incoming photon must be greater than or equal to the work function energy of the metal (3.61×10−19 J3.61 \times 10^{-19} \, \text{J}3.61×10−19J).

  • Red: 2.85×10−19 J2.85 \times 10^{-19} \, \text{J}2.85×10−19J (not enough energy)
  • Orange: 3.21×10−19 J3.21 \times 10^{-19} \, \text{J}3.21×10−19J (not enough energy)
  • Yellow: 3.38×10−19 J3.38 \times 10^{-19} \, \text{J}3.38×10−19J (not enough energy)
  • Green: 3.64×10−19 J3.64 \times 10^{-19} \, \text{J}3.64×10−19J (enough energy)
  • Blue: 3.98×10−19 J3.98 \times 10^{-19} \, \text{J}3.98×10−19J (enough energy)
  • Indigo: 4.11×10−19 J4.11 \times 10^{-19} \, \text{J}4.11×10−19J (enough energy)
  • Violet: 4.97×10−19 J4.97 \times 10^{-19} \, \text{J}4.97×10−19J (enough energy)

Conclusion

The colors of light that have enough energy to eject electrons from the metal are green, blue, indigo, and violet.

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