Chapter 5: Problem 52
The work function of cesium metal is \(188 \mathrm{~kJ} / \mathrm{mol}\), which corresponds to light with a wavelength of \(637 \mathrm{~nm}\). Which of the following will cause the smallest number of electrons to be ejected from cesium? (a) High-amplitude wave with a wavelength of \(500 \mathrm{~nm}\) (b) Low-amplitude wave with a wavelength of \(500 \mathrm{~nm}\) (c) High-amplitude wave with a wavelength of \(650 \mathrm{~nm}\) (d) Low-amplitude wave with a wavelength of \(650 \mathrm{~nm}\)
Short Answer
Step by step solution
Calculate Energy of the Incident Light
Determine Threshold Energy per Photon
Compare Energy Values to Work Function
Analyze Options
Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Work Function
Photon Energy Calculation
Wavelength and Frequency
Electron Ejection
- Photon Energy: Only photons with energy greater than the work function will cause ejection.
- Wavelength: Shorter wavelengths, corresponding to higher energies, are more effective in causing ejection.
- Intensity: Amplitude affects the number of photons but not the energy per photon.