What is the Maximum Wavelength of Light That Can Cause a Photoelectric Effect in Lithium?
Understanding the photoelectric effect is crucial in various fields, including quantum physics and material science. When light interacts with a material, it can displace electrons—it does so if the photon's energy exceeds the material's work function. This article explores the specific relationship between the maximum wavelength of light and the photoelectric effect in lithium, providing detailed steps and calculations.
The Physics Behind the Photoelectric Effect in Lithium
The photoelectric effect is a quantum electrodynamic phenomenon first detailed by Albert Einstein. It describes the ability of light to eject electrons from a material's surface. For the photoelectric effect to occur, the energy of the incident photon must exceed the material's work function, denoted as φ.
Author's Note: Li's Work Function and Its Implications
Lithium (Li) is a unique element with a work function of 2.5 electron volts (eV). To determine the maximum wavelength of light capable of causing the photoelectric effect, we need to understand the energy and wavelength relationship. This involves converting the work function from electron volts to joules, and then using the relationship between the energy of a photon and its wavelength.
Converting Work Function to Joules
First, let's convert lithium's work function from electron volts (eV) to joules (J). The conversion factor is 1 eV 1.602 x 10-19 J.
$$phi 2.5 , text{eV} times 1.602 times 10^{-19} , text{J/eV} 4.005 times 10^{-19} , text{J}$$
Calculating the Maximum Wavelength
The energy of a photon is given by the equation:
$$E frac{hc}{lambda}$$
where:
E is the energy of the photon (J) h is Planck's constant, h 6.626 x 10-34 Js c is the speed of light, c 3 x 108 m/s λ is the wavelength of the light (m)Setting the energy of the photon equal to the work function:
$$frac{hc}{lambda} phi$$
Rearranging this to solve for λ:
$$lambda frac{hc}{phi}$$
Substituting the values for h, c, and φ:
$$lambda frac{6.626 times 10^{-34} , text{Js} times 3 times 10^8 , text{m/s}}{4.005 times 10^{-19} , text{J}}$$
Calculating this gives:
$$lambda approx 4.96 times 10^{-7} , text{m} 496 , text{nm}$$
Therefore, the maximum wavelength of light that can cause the photoelectric effect in lithium is approximately 496 nm.
Alternative Method: Using the Photon Energy Formula
The alternative method to find the wavelength involves using the formula for photon energy:
$$E hf$$
where:
h is Planck's constant (6.626 x 10-34 Js) f is the frequency of the light (Hz) λ is the wavelength of the light (m) with the relationship between frequency and wavelength: λ c/fUsing the relation E φ and substituting:
$$hf frac{hc}{lambda}$$
$$lambda frac{h}{E/eV}$$
For E 3.6 eV:
$$lambda frac{6.626 times 10^{-34} times 3 times 10^8}{3.6 times 1.602 times 10^{-19}}$$
$$lambda approx 496 , text{nm}$$
Key Takeaways
The calculation confirms that a photon with a wavelength of approximately 496 nm or less has sufficient energy to cause the photoelectric effect in lithium. This wavelength falls within the ultraviolet (UV) range of the electromagnetic spectrum.
Related Concepts
The photoelectric effect is closely related to the understanding of the electromagnetic spectrum. Electromagnetic radiation is a form of energy that travels through space in the form of waves. In the case of lithium, the maximum wavelength for the photoelectric effect is approximately 496 nm, which is within the ultraviolet range.
Ultraviolet (UV) radiation has a wavelength ranging from 10 nm to 400 nm. The photoelectric effect in lithium is best seen with UV radiation, where the photon energy exceeds the work function of the material.