Math Problem Statement

A photon with a frequency of 5.23 E14 Hz strikes a photoemissive surface whose work function is 1.75 eV. Planck's constant is 4.14 E−15 eV*s.

Calculate the energy of the photon.

Calculate the maximum kinetic energy of the ejected photoelectron.

Calculate the threshhold frequency for the material.

Solution

Let's solve each part of the problem step-by-step.

1. Calculate the energy of the photon

The energy of a photon EE can be calculated using the equation: E=hfE = h \cdot f

Where:

  • hh is Planck's constant (4.14×10154.14 \times 10^{-15} eV·s)
  • ff is the frequency of the photon (5.23×10145.23 \times 10^{14} Hz)

Plugging in the values: E=4.14×1015 eV\cdotps×5.23×1014 HzE = 4.14 \times 10^{-15} \text{ eV·s} \times 5.23 \times 10^{14} \text{ Hz} E=2.17 eVE = 2.17 \text{ eV}

2. Calculate the maximum kinetic energy of the ejected photoelectron

The maximum kinetic energy KmaxK_{\text{max}} of the ejected photoelectron can be found using the photoelectric equation: Kmax=EϕK_{\text{max}} = E - \phi

Where:

  • EE is the energy of the photon (2.17 eV from the previous calculation)
  • ϕ\phi is the work function of the material (1.75 eV)

Plugging in the values: Kmax=2.17 eV1.75 eVK_{\text{max}} = 2.17 \text{ eV} - 1.75 \text{ eV} Kmax=0.42 eVK_{\text{max}} = 0.42 \text{ eV}

3. Calculate the threshold frequency for the material

The threshold frequency f0f_0 is the minimum frequency of light required to eject an electron from the material. It can be calculated using the work function ϕ\phi and Planck's constant hh: f0=ϕhf_0 = \frac{\phi}{h}

Where:

  • ϕ\phi is the work function (1.75 eV)
  • hh is Planck's constant (4.14×10154.14 \times 10^{-15} eV·s)

Plugging in the values: f0=1.75 eV4.14×1015 eV\cdotpsf_0 = \frac{1.75 \text{ eV}}{4.14 \times 10^{-15} \text{ eV·s}} f04.23×1014 Hzf_0 \approx 4.23 \times 10^{14} \text{ Hz}

Summary of Results

  1. Energy of the photon: 2.17 eV2.17 \text{ eV}
  2. Maximum kinetic energy of the ejected photoelectron: 0.42 eV0.42 \text{ eV}
  3. Threshold frequency for the material: 4.23×1014 Hz4.23 \times 10^{14} \text{ Hz}

Would you like more details or have any questions? Here are some related questions for further exploration:

  1. How does the energy of the photon change with frequency?
  2. What is the relationship between the work function and the threshold frequency?
  3. How can the kinetic energy of photoelectrons be measured experimentally?
  4. What happens to the kinetic energy of the photoelectrons if the frequency of the incoming photons is doubled?
  5. How does the material's work function affect the ejection of photoelectrons?
  6. What role does Planck's constant play in the photoelectric effect?
  7. How can the photoelectric equation be derived from basic principles?
  8. What other factors, besides frequency, can influence the energy of ejected photoelectrons?

Tip: Always ensure the units are consistent when using physical constants and equations to avoid calculation errors.

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Math Problem Analysis

Mathematical Concepts

Quantum Mechanics
Photoelectric Effect
Energy Calculations

Formulas

Photon Energy = Planck's constant * Frequency
Maximum Kinetic Energy = Photon Energy - Work Function
Threshold Frequency = Work Function / Planck's constant

Theorems

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Suitable Grade Level

Advanced High School or College