A beam of light with a frequency range of 3.01 x 1014 Hz - 6.10 x 1014 Hz is incident on a metal surface. If the work function of the metal surface is 2.20 eV, what is the range of frequencies in this beam of light that will be responsible for ejecting electrons from this metal surface? O 1.12 x 1014 Hz - 5.31 x 1014 Hz O 1.12 x 1014 Hz - 3.10 x 1014 Hz O None O 5.31 x 1014 Hz - 6.10 x 1014 Hz O 3.01 x 1014 Hz - 6.10 x 1014 Hz
A beam of light with a frequency range of 3.01 x 1014 Hz - 6.10 x 1014 Hz is incident on a metal surface. If the work function of the metal surface is 2.20 eV, what is the range of frequencies in this beam of light that will be responsible for ejecting electrons from this metal surface? O 1.12 x 1014 Hz - 5.31 x 1014 Hz O 1.12 x 1014 Hz - 3.10 x 1014 Hz O None O 5.31 x 1014 Hz - 6.10 x 1014 Hz O 3.01 x 1014 Hz - 6.10 x 1014 Hz
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![## Photoelectric Effect Problem
**Question:**
A beam of light with a frequency range of \(3.01 \times 10^{14} \, \text{Hz}\) - \(6.10 \times 10^{14} \, \text{Hz}\) is incident on a metal surface. If the work function of the metal surface is \(2.20 \, \text{eV}\), what is the range of frequencies in this beam of light that will be responsible for ejecting electrons from this metal surface?
**Options:**
1. \(1.12 \times 10^{14} \, \text{Hz} - 5.31 \times 10^{14} \, \text{Hz}\)
2. \(1.12 \times 10^{14} \, \text{Hz} - 3.10 \times 10^{14} \, \text{Hz}\)
3. None
4. \(5.31 \times 10^{14} \, \text{Hz} - 6.10 \times 10^{14} \, \text{Hz}\)
5. \(3.01 \times 10^{14} \, \text{Hz} - 6.10 \times 10^{14} \, \text{Hz}\)
### Explanation:
To solve this problem, one must understand the photoelectric effect. According to Einstein's photoelectric equation, electrons are ejected from a metal surface when the energy of the incident light (photons) exceeds the work function (\(\phi\)) of the metal. The energy of a photon is given by \(E = h \nu\), where \(h\) is Planck's constant and \(\nu\) is the frequency of the light.
The work function can be converted to frequency using:
\[ \phi = h \nu_0 \]
where \(\nu_0\) is the threshold frequency. For \(\phi = 2.20 \, \text{eV}\), and knowing that \(1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}\):
\[ \phi = 2.20 \times 1.602 \times 10^{-19} \, \text{J} = 3.5244 \times 10^{-19}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F185e23c6-dc6d-40cb-8ddc-039cdf270985%2Faef9d0f2-a699-4ee4-9a45-17bafdf89464%2Fgz9kn7t_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Photoelectric Effect Problem
**Question:**
A beam of light with a frequency range of \(3.01 \times 10^{14} \, \text{Hz}\) - \(6.10 \times 10^{14} \, \text{Hz}\) is incident on a metal surface. If the work function of the metal surface is \(2.20 \, \text{eV}\), what is the range of frequencies in this beam of light that will be responsible for ejecting electrons from this metal surface?
**Options:**
1. \(1.12 \times 10^{14} \, \text{Hz} - 5.31 \times 10^{14} \, \text{Hz}\)
2. \(1.12 \times 10^{14} \, \text{Hz} - 3.10 \times 10^{14} \, \text{Hz}\)
3. None
4. \(5.31 \times 10^{14} \, \text{Hz} - 6.10 \times 10^{14} \, \text{Hz}\)
5. \(3.01 \times 10^{14} \, \text{Hz} - 6.10 \times 10^{14} \, \text{Hz}\)
### Explanation:
To solve this problem, one must understand the photoelectric effect. According to Einstein's photoelectric equation, electrons are ejected from a metal surface when the energy of the incident light (photons) exceeds the work function (\(\phi\)) of the metal. The energy of a photon is given by \(E = h \nu\), where \(h\) is Planck's constant and \(\nu\) is the frequency of the light.
The work function can be converted to frequency using:
\[ \phi = h \nu_0 \]
where \(\nu_0\) is the threshold frequency. For \(\phi = 2.20 \, \text{eV}\), and knowing that \(1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}\):
\[ \phi = 2.20 \times 1.602 \times 10^{-19} \, \text{J} = 3.5244 \times 10^{-19}
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