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Chapter 27: Problem 11

Two different monochromatic light sources, one yellow \((580 \mathrm{nm})\) and one violet \((425 \mathrm{nm}),\) are used in a photoelectric effect experiment. The metal surface has a photoelectric threshold frequency of $6.20 \times 10^{14} \mathrm{Hz}$. (a) Are both sources able to eject photoelectrons from the metal? Explain. (b) How much energy is required to eject an electron from the metal? (Use \(h=4.136 \times\) \(10^{-15} \mathrm{eV} \cdot \mathrm{s} .\)

Short Answer

Expert verified
Based on the given information, determine if both yellow and violet light sources can produce the photoelectric effect on a metal surface with a threshold frequency of \(6.20 × 10^{14} \mathrm{Hz}\), and calculate the energy required to eject an electron from the metal. (a) For the yellow light source: Yes or No For the violet light source: Yes or No (b) The energy required to eject an electron from the metal: \(E_{threshold}\)

Step by step solution

01

1. Calculate the frequency of both light sources

We know that the speed of light (c) is around \(3 \times 10^8 \mathrm{m/s}\), and we can use the formula \(f = \frac{c}{\lambda}\) to calculate the frequency (f) of both light sources, where λ is the wavelength. For the yellow light source, the wavelength is \(580 \mathrm{nm}\), which is equal to \(5.8 \times 10^{-7}\mathrm{m}\). For the violet light source, the wavelength is \(425 \mathrm{nm}\), which is equal to \(4.25 \times 10^{-7}\mathrm{m}\).
02

2. Calculate the energy of both light sources

To calculate the energy of both light sources, we can use the formula \(E = h \cdot f\), where \(E\) is the energy, \(h\) is the Planck's constant (\(4.136 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s}\)), and \(f\) is the frequency from the previous step. For the yellow light source: \(E_{yellow} = h \cdot f_{yellow}\) For the violet light source: \(E_{violet} = h \cdot f_{violet}\)
03

3. Calculate the threshold energy of the metal

We are given the threshold frequency (\(f_{threshold} = 6.20 \times 10^{14} \mathrm{Hz}\)) to eject photoelectrons. We can find the threshold energy by using the same formula: \(E_{threshold} = h \cdot f_{threshold}\)
04

4. Determine if both light sources can eject photoelectrons

Compare the energies of both light sources to the threshold energy. If the energy of the light source is equal to or greater than the threshold energy, the light will eject photoelectrons. (a) For the yellow light source: If \(E_{yellow} \ge E_{threshold}\), it can eject photoelectrons. For the violet light source: If \(E_{violet} \ge E_{threshold}\), it can eject photoelectrons.
05

5. Calculate the energy required to eject an electron from the metal

(b) The energy required to eject an electron from the metal is the threshold energy, \(E_{threshold}\). We have calculated it in step 3 already.

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