Photoelectric Experiment Shows Violet Light Effect

A Photoelectric Experiment Indicates That Violet Light Of Wavelength 4

A photoelectric experiment indicates that violet light of wavelength 420 nm is the longest wavelength radiation that can cause photoemission of electrons from a particular multialkali photocathode surface. a. What is the work function in eV? b. If a UV radiation of wavelength 300 nm is incident upon the photocathode, what will be the maximum kinetic energy of the photoemitted electrons, in eV? c. Given that the UV light of wavelength 300 nm has an intensity of 20 mW/cm2, if the emitted electrons are collected by applying a positive bias to the opposite electrode, what will be the photoelectric current density in mA cm-2?

Paper For Above instruction

The photoelectric effect, first explained by Albert Einstein in 1905, fundamentally demonstrates how light interacts with matter, specifically how photons can eject electrons from a material surface when their energy exceeds the work function of the substance. This experiment involving violet light at a wavelength of 420 nm provides critical insights into the energy thresholds necessary for photoemission, and calculations based on this data allow us to determine the work function, analyze electron kinetic energies under different illumination conditions, and estimate current densities resulting from incident radiation. This paper explores these concepts through detailed thermodynamic and quantum mechanical calculations relevant to the photoelectric effect.

Introduction

The photoelectric effect is a cornerstone of quantum physics, illustrating the particle-like behavior of light. When photons with sufficient energy strike a material, they can eject electrons from the surface. The minimum energy required to liberate an electron is the work function, φ, inherent to the material. Einstein’s equation describes the relationship between photon energy, the work function, and the kinetic energy of emitted electrons:

KEmax = hf - φ

where h is Planck’s constant, f is the frequency of the incident photon, and KEmax is the maximum kinetic energy of the emitted electrons.

Part a: Determining the Work Function

The wavelength of the threshold radiation, λthreshold = 420 nm, corresponds to the longest wavelength capable of inducing photoemission. Converting this wavelength into energy using the relation E = hc/λ, where c is the speed of light and h is Planck’s constant:

E = (6.626×10-34 J·s)(3.0×108 m/s) / (420×10-9 m) ≈ 4.73×10-19 J

Since 1 eV = 1.602×10-19 J, the work function φ is:

φ = E / 1.602×10-19 ≈ (4.73×10-19 J) / (1.602×10-19 J/eV) ≈ 2.96 eV

Thus, the work function of the photocathode is approximately 2.96 eV.

Part b: Maximum Kinetic Energy of Photoelectrons for 300 nm UV Light

The energy of incident photons at a wavelength of 300 nm is calculated similarly:

E = (6.626×10-34 J·s)(3.0×108 m/s) / (300×10-9 m) ≈ 6.63×10-19 J

Converting to eV:

Ei = 6.63×10-19 J / 1.602×10-19 J/eV ≈ 4.14 eV

The maximum kinetic energy (KEmax) of the emitted electrons is the photon energy minus the work function:

KEmax = Ei - φ = 4.14 eV - 2.96 eV ≈ 1.18 eV

This indicates the electrons can acquire a maximum kinetic energy of approximately 1.18 eV when illuminated with 300 nm UV light.

Part c: Estimation of Photoelectric Current Density

The intensity (I) of the incident UV light is given as 20 mW/cm2. This relates to the photon flux (Φ) through the photon energy:

Φ = power / (area × photon energy) = I / Ephoton

Converting power to watts:

I = 20 mW/cm2 = 0.02 W/cm2

The photon flux per unit area is:

Φ = 0.02 W/cm2 / (4.14 eV × 1.602×10-19 J/eV) ≈ 0.02 / (6.63×10-19 J) ≈ 3.02×1016 photons/m2-s

(Here, 1 W = 1 J/s, and converting cm2 to m2 adds a factor of 104 in the denominator).

Assuming each incident photon ejects one electron and the quantum efficiency is maximum (close to 1), the current density (J) can be calculated as:

J = (electrons per second per area) × (charge of an electron)

Charge of an electron, e = 1.602×10-19 C

Therefore:

J = Φ × e ≈ 3.02×1016 electrons/m2-s × 1.602×10-19 C ≈ 4.84×10-3 C/m2-s

Converting to mA/cm2:

1 C/s = 1000 mA, and 1 m2 = 10,000 cm2:

J ≈ (4.84×10-3 C/m2-s) × (1000 mA / C) × (1 m2/10,000 cm2) ≈ 0.000484 mA/cm2

This current density is approximately 0.484 μA/cm2, which is substantially below 1 mA/cm2. If the quantum efficiency is less than 1, the actual current will be further reduced. Realistic efficiencies for multialkali photocathodes often range from 10% to 30%, which adjusts the current density proportionally upward within that range.

Thus, the estimated photoelectric current density under the given illumination conditions is approximately 0.5 μA/cm2, assuming near-unity quantum efficiency. Actual experimental measurements might vary depending on the material properties and surface conditions.

Conclusion

This analysis elucidates the fundamental parameters governing the photoelectric effect in the specified system. The work function is approximately 2.96 eV, setting the energy threshold for electron emission. Photons at 300 nm provide enough energy to generate electrons with maximum kinetic energy of about 1.18 eV. The photoelectric current density is modest under these conditions but can be enhanced by increasing intensity or quantum efficiency. These insights serve as a foundation for optimizing photoelectron emission in practical devices such as photomultiplier tubes and solar cells, highlighting the significance of photon energy, material properties, and illumination intensity in designing efficient photoelectric systems.

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